Math 10 Academic: MPM2D
Quadratic Functions: Basic Properties of Quadratic Relations
Unit
3: Quadratic Functions
Lesson
1: Basic Properties of Quadratic Relations
Expected Duration
Expectations
 graph any quadratic function in the form 𝑦 = 𝑎𝑥 2 using the Step Pattern;
 describe how the value of 'a' affects the orientation and shape of the parabola;
 identify the vertex and axis of symmetry for any parabola that can be written in the form
𝑦 = 𝑎𝑥 2 .
Introduction
Thus far we have been working with linear functions, the graphs of which are straight lines. In
contrast, the graphs of non-linear functions are curved. In this unit, we will be investigating
properties of quadratic functions. The graph of a quadratic function is in the shape of a
parabola. Examples of quadratic functions include the relationship between the height and
elapsed time for an object thrown upwards or an object in free fall. We can use quadratic
relationships to solve problems involving the motion of objects under the force of gravity, such
as a pebble that is dropped from a bridge into a river, a water droplet falling from a tree, or
the height of a flare as a function of time. Many archways are constructed in the shape of a
downwards-facing parabola to maximize their stability. The cables used in suspension bridges
tend to have the shape of a parabola. We will be investigating all of these applications, but first
we need to look at some basic properties of quadratic functions and their graphs.
Graph of 𝒚 = 𝒙𝟐
Let's consider the most basic quadratic function, 𝑦 = 𝑥 2 . We will complete a table for values
of x ranging from -5 to 5. The graph of 𝑦 = 𝑥 2 is shown to the right.
x
-4
-3
-2
-1
0
1
2
3
4
y
𝑦 = 𝑥2
𝑦 = (−4)2
𝑦 = 16
𝑦 = 𝑥2
𝑦 = (−3)2
𝑦=9
𝑦 = 𝑥2
𝑦 = (−2)2
𝑦=4
𝑦 = 𝑥2
𝑦 = (−1)2
𝑦=1
𝑦 = 𝑥2
𝑦 = 02
𝑦=0
𝑦 = 𝑥2
𝑦 = 12
𝑦=1
𝑦 = 𝑥2
𝑦 = 22
𝑦=4
𝑦 = 𝑥2
𝑦 = 32
𝑦=9
𝑦 = 𝑥2
𝑦 = 42
𝑦 = 16
Ordered
Pair
(-4, 16)
(-3, 9)
(-2, 4)
(-1, 1)
(0, 0)
(1, 1)
(2, 4)
(3, 9)
(4, 16)
The graph of 𝑦 = 𝑥 2 has these properties.
The graph is a parabola opening upward.
The origin is the lowest point of the parabola.
The graph is symmetrical about the y axis.
The first differences are the odd numbers 1, 3, 5, 7, , 11, etc.
Now let's look at the differences in the y coordinates for x values ranging from 0 to 6. These
are called the first differences. We will be able to use them to graph the function 𝑦 = 𝑥 2 .
x
y
0
0
1
1
2
4
3
9
4
16
5
25
6
36
First
Differences
Graphing 𝑦 =
𝑥2
Plot (0, 0)
1−0=1
Go 1 right and 1
up to (1, 1)
4−1=3
Go 1 right and 3
up to (2, 4)
9−4=5
Go 1 right and 5
up to (3, 9)
16 − 9 = 7
Go 1 right and 7
up to (4, 16)
25 − 16 = 9 Go 1 right and 9
up to (5, 25)
36 − 25 = 11 Go 1 right and
11 up to (6, 36)
As described in the last column in the table
shown above, we can use the first differences
to graph 𝑦 = 𝑥 2 . We start by plotting the
origin (0, 0). Then we move 1 right and 1 up,
then 1 right and 3 up, then 1 right and 5 up,
then 1 right and 7 up, and so on Points on
the left side of the origin are found by
symmetry. Then we draw a smooth curve
through the points.
The graph of 𝒚 = 𝒙𝟐 has the following properties:
 The graph is a parabola opening upward.
 The origin is the lowest point of the parabola. This point is also called the vertex of the
parabola.
 The graph is symmetrical about the y axis. The y axis is the axis of symmetry for the
graph.
Example 1
Use the step pattern to graph the function 𝑦 = 0.5𝑥 2 . Also use the Step Pattern to graph
𝑦 = 2𝑥 2 . Graph the function 𝑦 = 𝑥 2 on the same grid. Explain how the shape of the parabola
changes as the value of 'a' changes.
Solution
The Step Pattern for 𝑦 = 𝑥 2 is 1 right and 1 up, 1 right and 3 up, 1 right and 5 up,...
We notice from the graphs shown above that:
(a) The graph of 𝑦 = 2𝑥 2 is narrower or vertically stretched as compared to the graph of
𝑦 = 𝑥 2.
(b) The graph of 𝑦 = 0.5𝑥 2 is wider or vertically compressed as compared to the graph of
𝑦 = 𝑥 2.
When the value of a is between -1 and 1, the parabola is vertically compressed
as compared to the graph of 𝒚 = 𝒙𝟐 . When the value of a is less than -1 or greater
than 1, the parabola is vertically stretched as compared to the graph of 𝒚 = 𝒙𝟐 .
For these reasons, a is called the stretch factor.
Let's look at an example of a parabola for which the stretch factor a is negative.
Example 2
Graph the function 𝑦 = −3𝑥 2 .
Solution
The Step Pattern for 𝑦 = 𝑥 2 is 1, 3, 5, 7,...
The Step Pattern for 𝑦 = −3𝑥 2 will be
−3 × 1 = −𝟑, −3 × 3 = −𝟗,
−3 × 5 = −𝟏𝟓, −3 × 7 = −𝟐𝟏, …
Since the steps for this graph are negative,
we move down at each step.
To graph 𝑦 = −3𝑥 2 then, we plot the
vertex (0, 0) and apply the Step Pattern:
1 right and 3 down, 1 right and 9 down,
Comparing Quadratic Functions to Linear Functions
Recall that linear functions have constant first differences. Let's look at the first differences
for the linear function 𝑦 = 2𝑥 + 1.
𝑥
𝑦 = 2𝑥 + 1
First Differences
-2 𝑦 = 2(−2) + 1
𝑦 = −4 + 1
𝑦 = −𝟑
-1 𝑦 = 2(−1) + 1 −𝟏 − (−𝟑)
𝑦 = −2 + 1
= −1 + 3)
𝑦 = −𝟏
=2
0
𝑦 = 2(0) + 1
𝑦 =0+1
𝑦=1
1 − (−1)
= 1 + 1)
=2
1
𝑦 = 2(1) + 1
𝑦 =2+1
𝑦=3
3−1=2
Now let's look at the first and second differences for the quadratic function 𝑦 = 𝑥 2 .
𝑥
𝑦 = 𝑥2
First Differences Second Differences
-3 𝑦 = (−3)2
𝑦=9
-2 𝑦 = (−2)2 4 − 9 = −𝟓
𝑦=4
-1 𝑦 = (−1)2 1 − 4 = −𝟑
𝑦=1
−𝟑 − (−𝟓)
= −3 + 5
=2
0
𝑦 = 02
𝑦=0
0 − 1 = −𝟏
−𝟏 − (−𝟑)
= −1 + 3
=2
1
𝑦 = 12
𝑦=1
1−0=1
1 − (−1)
=1+1
=2
2
𝑦 = 22
𝑦=4
4−1=3
3−1=2
2
Assignment 1: Basic Properties of Quadratic Relations
1. a) Complete the table for the points shown on the parabola below.
𝑥 𝑦
-3 -18
-2
-8
-1
-2
0
0
1
-2
2
-8
3
-18
First
Second
Differences Differences
Assignment 1: Answers
1. (a)
𝑥 𝑦
-3 -18
1
-2
-8
10
-1
-2
6
-4
0
0
2
-4
1
-2
-2
-4
2
-8
-6
-4
3
-18
-10
-4
b) 𝑎 = (−4) = −2
2
2
First
Second
Differences Differences
Math 10 Academic: MPM2D
Quadratic Functions: The Parabola of Best Fit
Unit
3: Quadratic Functions
Lesson
2: The Parabola of Best Fit
Expected Duration
Expectations




collect and graph data that can be represented as a quadratic relation;
use technology to draw a curve of best fit;
use technology to determine the equation of the parabola of best fit;
apply a model to solve problems involving projectile motion
Introduction
Using Quadratic Functions to Model Parabolic Structures
The Gateway Arch
When the Gateway Arch, shown above, is placed on an x-y grid with its highest point at the
in St. Louis,
Missouri
origin, the following orderedGateway
pairs areArch
obtained.
For each
point, the first coordinate is the
horizontal distance in metres from the vertex and the second coordinate is the vertical
distance, also in metres, from the vertex.
(0, 0), (12, -1.2), (24, -4.8), (36, -14.4), (48, -26.4), (60, -45.6), (72, -72), (84, -112.8),
(96, -176.4), (-12, -1.2), (-24, -4.8), (-36, -14.4), (-48, -26.4), (-60, -45.6), (-72, -72),
As is the case for most structures, the curve of best fit for the Gateway Arch is not perfectly
modelled by a parabola. In order for the highest point of the arch to correspond to the vertex
of the parabola, the two points to the far right and the far left of the scatter plot had to be
removed from the calculation. What the curve of best fit reveals to us however is that the top
half of the Gateway Arch is very closely modelled by a parabola. The base of the structure,
however, deviates from the parabolic shape and has a narrower width than the graph of 𝑦 =
2
y
Suspension Bridges
This suspension bridge is the Akashi-Kaikyo Suspension Bridge in Kobe, Japan. Later in this unit,
we will use mathematical methods to develop an equation that models the shape of this
bridge. In this section, we'll use technology to show that suspension bridges can be modeled
quite accurately by quadratic functions.
The following diagram illustrates all the components of a suspension bridge. A suspension
bridge can be modelled using a quadratic function. As was done for the Gateway Arch, we
place the bridge on a pair of x and y axes and collect position data. You can verify the ordered
pairs that are obtained:
(0, 5.5), (2, 5.7), (4, 6.3), (6, 7.3), (8, 9), (10, 12), (-2, 5.7), (-4, 6.3), (-6, 7.3), (-8, 9), (-10, 12)
Plotting this data in Geogebra gives us a scatter plot. Superimposed on the scatter plot is the
parabola of best fit.
Projectile Motion
A stroboscopic picture is one in which the object has been photographed at specific time
intervals during its motion. We are going to investigate the data obtained from a stroboscopic
picture of a ball thrown almost straight upwards. First we will display the data points in a
scatter plot and then find the equation for the parabola of best fit. The stroboscope flashes
every 0.1 seconds and records the image of the ball in midflight. The first coordinate in each
ordered pair is the time passed since the ball was thrown in seconds and the second coordinate
Height of a Projectile
A. Projected Vertically Upward
The height h, in metres, of an object projected vertically upward is given by the formula
ℎ = −4.9𝑡 2 + 𝑢𝑡
where t, in seconds, is the elapsed time, and u is the initial vertical speed in metres per second.
Assignment 2: Working with Quadratic Models
1. In a science experiment, water is allowed to drip steadily
from a burette. The falling droplets of water appear to be
stopped in this stroboscopic picture. The scale of the picture
is 1 cm represents 0.1 m. The time between successive flashes
of the stroboscope is 0.1 seconds. The data consists of the
elapsed time and the vertical distance relative to the white line
marked on the picture. The vertical distance of the white line is
chosen to be ℎ = 0. After the data is collected, it is recorded in
a table as shown below. The first flash was taken at a time of
0 seconds and a vertical distance from the tip of the burette of
0 metres.
Image Elapsed Time
Number
(seconds)
1
0
2
0.1
3
0.2
4
0.3
5
0.4
Vertical distance Actual Vertical
as measured on
Distance (m)
the picture (cm)
1 cm = 0.1 m
0
0
-0.5
-0.5 × 0.1 = -0.05
-2.0
-4.4
-7.8
a) Complete the table.
b) Make a scatter plot of the Actual Vertical Distance (in metres) against the Elapsed Time
(in seconds). The first ordered pair is (0, 0). Use technology to determine the
equation of the parabola of best fit.
c) Compare the equation of the parabola of best fit to the equation for the height of an
object in free fall. How accurate is your equation?
2. Suppose a slingshot is used to throw a stone vertically upward at 40 m/s.
a) Substitute 40 in for 𝑢 in the equation ℎ = −4.9𝑡 2 + 𝑢𝑡 to express the height of the stone
as a function of the elapsed time. Graph the function.
b) Use your graph to estimate the maximum height of the stone, and the time it takes to
reach the maximum height.
c) Use your graph to estimate the time it takes for the stone to return to the ground.
Assignment 2: Answers
1. a)
Image Elapsed Time
Number
(seconds)
1
0
2
0.1
3
0.2
4
0.3
5
0.4
Vertical distance
as measured on
the picture (cm)
0
-0.5
-2.0
-4.4
-7.8
Actual Vertical
Distance (m)
1 cm = 0.1 m
0
-0.05
-0.20
-0.44
-0.78
𝑏) ℎ = −4.8𝑡 2 − 0.04𝑡
𝑐) ℎ = −4.8𝑡 2 − 0.04𝑡 is reasonably close to ℎ = −4.9𝑡 2 , within experimental error
caused by the limitations of the stroboscope.
2. 𝑎) ℎ = −4.9𝑡 2 + 40𝑡
b) maximum height is 82 metres at a time of 4 seconds
c) 8 seconds
Math 10 Academic: MPM2D
Quadratic Functions: Comparing Quadratic Functions and Quadratic Equations
Unit
Lesson
Expected
Duration
3: Quadratic Functions
3: Graphs of Quadratic Functions
Expectations
 describe the differences between quadratic functions and quadratic equations;
 use technology to identify the properties of quadratic functions;
 use technology solve quadratic equations
Introduction
In Lesson 1 and 2, we investigated a number of examples of quadratic functions and explored
how a quadratic model can be used to solve problems involving the physics of motion. In
Lesson 1, a quadratic function was defined as a function that can be written in the form 𝑦 =
𝑎𝑥 2 + 𝑏𝑥 + 𝑐. We explored how the shape of the graph is influenced by the value of 'a'. We
also discussed how to use the vertex and the step pattern to construct the graph of the
function.
As we have seen in Lessons 1 and 2, when a quadratic function is written in the form 𝑦 =
𝑎𝑥 2 + 𝑏𝑥 + 𝑐, called Standard Form, its vertex is not the origin (0, 0). In order to determine
the vertex of such a function, we either need to use technology to graph the function or we
need to write the function in a different form that would reveal its vertex to us. Later on in this
unit we will discuss how to express a quadratic function in vertex form. When a quadratic
function is in vertex form, its vertex is readily apparent and we can graph the function using its
vertex and step pattern just as we did for each function in Lesson 1. In this lesson, we will use
technology to graph quadratic functions and determine important properties of the function
from the graph, including the vertex, the y-intercepts, and the x-intercepts. We will also show
that the x-intercepts of a quadratic function that is in the form
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 are the solutions to the corresponding equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0.
Example 1
a) Graph the function 𝑦 = 𝑥 2 − 2𝑥 − 8.
b) Determine the y-intercept and state the corresponding ordered pair. How could the yintercept be determined algebraically?
c) Determine the x-intercepts and state the corresponding ordered pairs.
d) What is the value of the y coordinate for any x-intercept? State the quadratic equation
that is satisfied by these x-intercepts.
e) Write the equation for the axis of symmetry.
f) How could we have obtained the equation for the axis of symmetry from the xintercepts?
g) Determine the coordinates of the vertex. Is the vertex a minimum or maximum?
h) How could we have algebraically determined the y coordinate of the vertex from the
equation for the axis of symmetry?
Solution
a)
b) From the graph, we see that the y-intercept is -8. The ordered pair is (0, -8). We can also
calculate the value of the y-intercept by substituting 0 in for x in the function.
𝑦 = 𝑥 2 − 2𝑥 − 8.
𝑦 = 02 − 2(0) − 8
𝑦 = −8
This is consistent with the y-intercept found by graphing.
c) The x-intercepts are -2 and 4. The ordered pairs are (-2, 0) and (4, 0).
d) For any x-intercept, 𝑦 = 0. The quadratic equation that is satisfied by the x-intercepts -2
and 4 is determined by substituting 0 in for y in the quadratic function 𝑦 = 𝑥 2 − 2𝑥 − 8.
The equation is therefore 𝑥 2 − 2𝑥 − 8 = 0. The x-intercepts of the parabola defined by
𝑦 = 𝑥 2 − 2𝑥 − 8 are also called the roots of the quadratic equation 𝑥 2 − 2𝑥 − 8 = 0.
e) The equation of the axis of symmetry is 𝑥 = 1.
f) The axis of symmetry is located half way between the x-intercepts -2 and 4. Therefore we
can algebraically determine the equation of the axis of symmetry by finding the midpoint
of the x-intercepts.
−2 + 4
𝑥=
2
2
𝑥=
2
𝑥=1
Therefore the equation of the axis of symmetry is 𝑥 = 1.
g) The coordinates of the vertex are (1, -9). Since the parabola opens upwards, the vertex is
a minimum. The minimum value of the function is the y coordinate of the vertex -9.
h) In the equation of the axis of symmetry, 𝑥 is equal to the x coordinate of the vertex, 1.
To algebraically determine the y coordinate of the vertex, we need to the x value of 1
into the function 𝑦 = 𝑥 2 − 2𝑥 − 8.
𝑦 = 12 − 2(1) − 8
𝑦=1−2−8
𝑦 = −9
This is consistent with the graph that shows the vertex is (1, -9).
Example 2
a) Solve the quadratic equation 2𝑥 2 + 9𝑥 − 18 = 0 by graphing.
b) Determine the equation of the axis of symmetry for the graph of 𝑦 = 2𝑥 2 + 9𝑥 − 18
by calculating the midpoint of the roots of the equation 2𝑥 2 + 9𝑥 − 18 = 0. Compare
your result to the axis of symmetry shown on the graph.
c) The x coordinate of the vertex can also be calculated by using the formula
𝑏
𝑥=−
2𝑎
where the values of a and b are obtained from the first two coefficients of the function
𝑦 = 2𝑥 2 + 9𝑥 − 18. Use this formula to calculate the x coordinate of the vertex and
compare your result to the equation of the axis of symmetry found in part (b).
d) Calculate the y coordinate of the vertex by using your result in part (b). State the
coordinates of the vertex and compare your result to the vertex on the graph.
Solution
a) The roots of the equation 2𝑥 2 + 9𝑥 − 18 = 0 are the x-intercepts of the function
𝑦 = 2𝑥 2 + 9𝑥 − 18. We need to graph this function. From the graph below, we see that
the roots are -6 and 1.5. We also notice that the parabola opens upwards and the vertex
is a minimum.
b) The roots were found to be -6 and 1.5. The midpoint or mean value of these roots will
give us the x coordinate of the vertex and the equation of the axis of symmetry.
−6 + 1.5
2
−4.5
𝑥=
2
𝑥=
𝑥 = −2.25
Therefore the equation of the axis of symmetry is 𝑥 = −2.25 and the x coordinate of the
vertex is -2.25. This is consistent with the graph of the function 𝑦 = 2𝑥 2 + 9𝑥 − 18.
c) The x coordinate of the vertex and the axis of symmetry can also be found by using the
formula 𝑥 = −
𝑏
2𝑎
. For the function 𝑦 = 2𝑥 2 + 9𝑥 − 18, 𝑎 = 2 and 𝑏 = 9.
𝑏
2𝑎
9
𝑥=−
2(2)
𝑥=−
𝑥=−
9
4
𝑥 = −2.25
This is consistent with the equation of the axis of symmetry found in part (b).
d) The x coordinate of the vertex is -2.25. We can calculate the y coordinate of the vertex by
substituting -2.25 into the x variable in the function 𝑦 = 2𝑥 2 + 9𝑥 − 18.
𝑦 = 2(−2.25)2 + 9(−2.25) − 18
𝑦 = 10.125 − 20.25 − 18
𝑦 = −28.125
The y coordinate of the vertex is therefore -28.125 and the corresponding ordered pair is
(-2.25, -28.125). This result, determined algebraically, is consistent with the graph of the
function 𝑦 = 2𝑥 2 + 9𝑥 − 18. Since the vertex is a minimum, we say that the minimum
value of the function 𝑦 = 2𝑥 2 + 9𝑥 − 18 is the y coordinate of the vertex -28.125.
Summary of the Properties of Quadratic Functions
 A quadratic equation is an equation that can be written in the form 𝒚 =
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, where a, b, and c are constants and a≠0.
 When the value of 'a' is between -1 and 1, the parabola is vertically
compressed as compared to the graph of 𝒚 = 𝒙𝟐 .
 When the value of 'a' is less than -1 or greater than 1, the parabola is
vertically stretched as compared to the graph of 𝒚 = 𝒙𝟐 .
 When 'a' is positive, the parabola opens upwards and the vertex of the
parabola is the lowest point, or minimum. The y coordinate of the vertex
is the minimum value of the function.
 When 'a' is negative, the parabola opens downwards and the vertex of the
parabola is the highest point, or maximum. The y coordinate of the vertex
is the maximum value of the function.
 The y-intercept of a parabola is the value of y at which the parabola
intersects the y axis.
 The y-intercept can be found algebraically by substituting 0 in for x in the
function and calculating the value of y.
 The x-intercepts of a parabola are the values of x at which the parabola
intersects the x axis.
 The x-intercepts of a parabola defined by 𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 are also
called the roots of the quadratic equation 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎.
 The axis of symmetry is a vertical line that passes through the vertex.
 The equation of the axis of symmetry and the x coordinate of the vertex is
calculated by either (a) finding the midpoint of the x-intercepts or
(b) using the formula 𝒙 = −
𝒃
𝟐𝒂
where a and b are
found from the function 𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄.
 The y coordinate of the vertex can be determined by substituting the x
coordinate of the vertex into the quadratic function.
Assignment 3: Graphs of Quadratic Functions
1. a) Solve the quadratic equation 𝑥 2 − 𝑥 − 6 = 0 by graphing.
b) Determine the equation of the axis of symmetry for the graph of 𝑦 = 𝑥 2 − 𝑥 − 6
by calculating the midpoint of the roots of the equation 𝑥 2 − 𝑥 − 6 = 0. Compare
your result to the axis of symmetry shown on the graph.
c) Use the formula 𝑥 = −
𝑏
2𝑎
to determine the equation of the axis of symmetry
and the x coordinate of the vertex. Compare your result to part (b).
d) Is the vertex a minimum or maximum?
e) Calculate the minimum value of the function (the y coordinate of the vertex) by
substituting the x coordinate of the vertex into the function. State the vertex and
compare your result to the graph.
2. a) Graph the parabola described by 𝑦 = −𝑥 2 + 7𝑥 + 8 and use the graph to solve the
quadratic equation−𝑥 2 + 7𝑥 + 8 = 0.
b) Multiply every term in the equation −𝑥 2 + 7𝑥 + 8 = 0 by −1 to show the equation
is equivalent to 𝑥 2 − 7𝑥 − 8 = 0.
c) Determine the equation of the axis of symmetry and the x coordinate of the vertex
using both methods from 1(b) and 1(c) and compare your results to the graph.
d) Is the vertex of the parabola 𝑦 = −𝑥 2 + 7𝑥 + 8 a maximum or minimum?
e) Calculate the maximum value of the function (the y coordinate of the vertex) by
making the appropriate substitution. State the vertex and compare your result to the
graph.
Assignment 3: Answers
1. a) roots are 3 and -2
b) 𝑥 = 0.5
c) 𝑥 = 0.5
d) minimum
e) minimum value is −6.25, vertex is (0.5, −6.25)
2. a) roots are 8 and -1
b) (−1)(−𝑥 2 ) + (−1)(7𝑥) + (−1)(8) = 0
𝑥 2 − 7𝑥 − 8 = 0
c) 𝑥 = 3.5
d) maximum
e) maximum value is 20.25, vertex is (3.5, 20.25)
Math 10 Academic: MPM2D
Quadratic Functions: Common Factoring
Unit
3: Quadratic Functions
Lesson
4: Common Factoring
Expected Duration
Expectations
 use common factoring to solve quadratic equations.
Introduction
In Lesson 3, we solved a quadratic equation graphically by using technology to find the xintercepts of the corresponding quadratic equation. From this point on in the unit, we will
develop algebraic strategies to solve quadratic equations. Some quadratic equations can be
solved by using common factoring. In this lesson, we will develop the method of common
factoring and use it to solve quadratic equations.
Common Factoring
To factor an integer means to write it as a product of other integers. For example, 36 can be
factored in different ways:
36 = 2 × 18
36 = 4 × 9
36 = 6 × 6
36 = 2 × 2 × 9
36 = 2 × 3 × 6
36 = 2 × 2 × 3 × 3
When we write an integer as a product of its prime factors, we say that the integer has been
factored fully. (A prime number or prime factor has only two factors; itself and 1.) When we
write 36 = 2 × 2 × 3 × 3, we have fully factored 36, since 2 and 3 are both prime factors.
To factor a polynomial also means to express it as a product of other factors. However when
we factor a polynomial, we need to find the greatest common factor of all the terms in the
polynomial. The greatest common factor is the factor that is common to all the terms and is
the only factor that will fully factor the polynomial.
Example 1
Factor each binomial. (A binomial is a polynomial that has two terms.)
𝑎) 3𝑥 + 9
𝑏) 8 − 12𝑎
𝑐) 15𝑥 + 5𝑥 2
𝑑) 12𝑎𝑏 2 − 30𝑎2 𝑏
circle
Solution
a) We need to find the greatest common factor of each term; 3𝑥 and 9. Since the variable 𝑥
is only part of the first term, it is not common to both terms, so the greatest common
factor (also called the GCF) will be the GCF of the integers 3 and 9. The GCF of 3 and 9 is 3.
Now when we factor the GCF out of each term, the second factor is found by dividing
each term by the GCF. The factored form of the binomial 3𝑥 + 9 is as follows.
3𝑥 + 9 = 3 (
3𝑥 9
+ )
3 3
= 3(𝑥 + 3)
b) We need to find the greatest common factor of each term; 8 and 12𝑎. Since the variable
𝑎 is only part of the second term, it is not common to both terms, so the greatest common
factor will be the GCF of the integers 8 and 12. The GCF of 8 and 12 is 4. As in (a), when we
factor the GCF out of each term, the second factor is found by dividing each term by the
GCF. The factored form of the binomial 8 − 12𝑎 is as follows.
8 12𝑎
8 − 12𝑎 = 4 ( −
)
4
4
= 4(2 − 3𝑎)
c) We need to find the greatest common factor of each term; 15𝑥 and 5𝑥 2 . Since the
variable 𝑥 is present in both terms, the greatest common factor will consist of an integer
and a variable in this case. First we need to find the greatest common factor of the
integers 15 and 5. That is 5.
When every term in a polynomial has the same variable (𝑥 in this case), but raised to
different powers, the greatest common factor is the lowest power to which the variable
is raised. In the first term, 15𝑥, the variable 𝑥 is raised to the power 1 (𝑥 = 𝑥 1 ). In the
second term, 5𝑥 2 , the variable 𝑥 is raised to the power 2. Therefore the lowest power is
1 and we factor 𝑥 out of each term along with the integer 5. The greatest common factor
then is 𝟓𝒙. As in (a) and (b), when we factor out the GCF, we obtain the second factor
by dividing each term by the GCF.
15𝑥 5𝑥 2
15𝑥 + 5𝑥 = 5𝑥 (
+
)
5𝑥
5𝑥
2
= 5𝑥(3 + 𝑥)
𝑾𝒉𝒆𝒏 𝒘𝒆 𝒅𝒊𝒗𝒊𝒅𝒆 𝒑𝒐𝒘𝒆𝒓𝒔 𝒐𝒇 𝒙, 𝒘𝒆 𝒔𝒖𝒃𝒕𝒓𝒂𝒄𝒕
𝒙𝟐
𝒕𝒉𝒆 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒕𝒔. 𝑺𝒐
= 𝒙𝟐−𝟏 = 𝒙𝟏 = 𝒙
𝒙
d) We need to find the greatest common factor of each term; 12𝑎𝑏 2 and 30𝑎2 𝑏 . The
greatest common factor will consist of an integer and two variables here. The greatest
common factor of the integers 12 and 30 is 6.
Now we need to find the greatest common factor of the variables 𝑎 and 𝑏. Again, for
every variable in an expression, the greatest common factor is the lowest power to
which the variable is raised. In the first term, 12𝑎𝑏 2 , the variable 𝑎 is raised to the
power 1 and the variable 𝑏 is raised to the power 2. In the second term, 30𝑎2 𝑏 , the
variable 𝑎 is raised to the power 2 and the variable 𝑏 is raised to the power 1. Therefore
the lowest power to which 𝑎 𝐚𝐧𝐝 𝑏 are both raised is 1 and we factor 𝑎𝑏 out of each
term along with the integer 6. The greatest common factor then is 𝟔𝒂𝒃 and the second
factor of the factored form of the binomial 12𝑎𝑏 2 − 30𝑎2 𝑏 is found by dividing each term
by the GCF.
12𝑎𝑏 2 30𝑎2 𝑏
𝒃𝟐
𝒂𝟐
2
2
𝟐−𝟏
12𝑎𝑏 − 30𝑎 𝑏 = 6𝑎𝑏 (
−
=𝒃
= 𝒃 𝐚𝐧𝐝
= 𝒂𝟐−𝟏 = 𝒂
)
6𝑎𝑏
6𝑎𝑏
𝒃
𝒂
= 6𝑎𝑏(2𝑏 − 5𝑎)
Example 2
Factor each trinomial. (A trinomial is a polynomial that has three terms.)
𝑎) 25𝑏 2 − 30𝑏 + 10
𝑏) 57𝑝𝑞 4 + 27𝑝𝑞 3 − 9𝑝𝑞 2
Solution
a) Since the variable 𝑏 is only present in two of the terms, the greatest common factor will
be the GCF of the integers 25, 30, and 10. Therefore the GCF is 5. We factor out the GCF
and obtain the second factor by dividing each term by the GCF in brackets.
25𝑏2 30𝑏 10
25𝑏 − 30𝑏 + 10 = 5 (
−
+ )
5
5
5
2
= 5(5𝑏2 − 6𝑏 + 2)
b) Since both variable 𝑝 and 𝑞 are present in every term, the greatest common factor will
consist of an integer, a power of 𝑝, and a power of 𝑞. The GCF of the integers 57, 27, and
9 is 3. Since the lowest power of 𝑝 in the expression 57𝑝𝑞 4 + 27𝑝𝑞 3 − 9𝑝𝑞 2 is 1, we
factor out 𝑝. Since the lowest power of 𝑞 in the expression is 2, we also factor out 𝑞 2 .
Therefore the GCF is 𝟑𝒑𝒒𝟐 . We factor out the GCF and obtain the second factor by
dividing each term by the GCF in brackets.
57𝑝𝑞 4 27𝑝𝑞 3 9𝑝𝑞 2
57𝑝𝑞 + 27𝑝𝑞 − 9𝑝𝑞 = 3𝑝𝑞 (
+
−
)
3𝑝𝑞 2
3𝑝𝑞 2
3𝑝𝑞 2
4
3
2
2
2(
2
= 3𝑝𝑞 19𝑞 + 9𝑞 − 3)
𝒒𝟒
= 𝒒𝟒−𝟐 = 𝒒𝟐 𝒂𝒏𝒅
𝟐
𝒒
𝒒𝟑
= 𝒒𝟑−𝟐 = 𝒒
𝟐
𝒒
Example 2
Solve each of the equations.
𝑏) 8𝑦 2 − 12𝑦 = 0
𝑎) 15𝑥 − 30 = 0
Solution
We need to factor out the GCF in each case.
𝑎) 15𝑥 − 30𝑥 2 = 0
15𝑥(𝑥 − 2) = 0
Now we can divide both sides by 15 to simplify the solution.
15𝑥(𝑥 − 2)
0
=
15
15
𝑥(𝑥 − 2) = 0
Now we have an equation in which the product of two factors 𝑥 and x − 2 is equal
to zero. In order for this to be true, either the first factor 𝑥 is zero or the second factor
𝑥 − 2 is zero or both factors are zero. To solve for the two roots, we assume both factors
are zero.
Therefore 𝑥 = 0
and
The two roots are 0 and 2.
𝑥−2=0
𝑥=2
𝑏) 8𝑎2 − 12𝑎 = 0
4𝑎(2𝑎 − 3) = 0
Now we can divide both sides by 4 to simplify the solution.
4𝑎(2𝑎 − 3) 0
=
4
4
𝑎(2𝑎 − 3) = 0
As on (a), we assume both factors are zero.
Therefore 𝑎 = 0
and
2a − 3 = 0
2𝑎 = 3
2𝑎 3
=
2
2
3
𝑎=
2
3
The two roots are 0 and .
2
Example 3
Recall the second question in Assignment 2 described a slingshot that was used to throw a
stone vertically upward at 40 m/s. The equation that describes the motion of this projectile
can be approximated by ℎ = −5𝑡 2 + 40𝑡.
a) Use common factoring to determine the time it takes for the stone to return to the
ground.
b) Determine the coordinates of the vertex. What is the maximum height of the stone and
the time it took to reach that maximum height?
Solution
a) When the stone returns on the ground, the height ℎ = 0. To find the times at which the
stone is on the ground, we need to substitute 0 in for the height and solve the quadratic
equation for the two roots.
0 = −5𝑡 2 + 40𝑡
Since we need to find the roots of this equation, it is preferable to switch the sides.
−5𝑡 2 + 40𝑡 = 0
Now we factor out the greatest common factor. The GCF of the integers -5 and 40 is -5.
(Since the coefficient of 𝑡 2 is negative, we factor out -5. This simplifies the solution.)
The lowest power of 𝑡 is 1, so we also factor out 𝑡. Therefore the GCF is −5𝑡. We factor
−5𝑡 out of each term on the left side of the equation.
−5𝑡 2 40𝑡
−5𝑡 (
+
)=0
−5𝑡 −5𝑡
−5𝑡(𝑡 − 8) = 0
We divide both sides of the equation by -5 to simplify the solution.
−5𝑡(𝑡 − 8)
0
=
−5
−5
𝑡(𝑡 − 8) = 0
To solve for the two roots, we assume both factors are zero.
Therefore 𝑡 = 0
and
𝑡−8=0
𝑡=8
The times at which ℎ = 0 and the stone is on the ground are 0 seconds and 8 seconds.
These times are the x-intercepts of the parabola. The time of 0 seconds corresponds to
the time at which the stone was launched from ground level into the air. The time of 8
seconds is the time at which the stone returned to the ground.
b) To find the x coordinate of the vertex, we need to calculate the midpoint of the
x-intercepts 0 and 8.
𝑥𝑣𝑒𝑟𝑡𝑒𝑥 =
0+8 8
= =4
2
2
We determine the maximum height reached by the stone by substituting 𝑥𝑣𝑒𝑟𝑡𝑒𝑥 = 4 in
for the time in the original function ℎ = −5𝑡 2 + 40𝑡.
ℎ = −5(4)2 + 40(4)
ℎ = −80 + 160
ℎ = 80
Therefore the maximum height reached by the stone was 80 metres. This height was
reached at a time of 8 seconds after the stone was launched into the air.
Assignment 4: Common Factoring
1. Factor each of the following.
𝑎) 6𝑎 + 18
𝑏) 3𝑥 2 − 9𝑥
𝑐) 36𝑐 − 45𝑐 2
𝑑) 15𝑝2 − 55𝑝
𝑒) 2𝑑2 + 4𝑑 + 2
𝑓) 16𝑥 2 − 32𝑥 − 8
𝑔) 9𝑥 2 + 6𝑥 − 12
ℎ) 24𝑑2 + 40𝑑 + 56
2. Solve each equation.
𝑎) 2𝑎2 − 6𝑎 = 0
𝑏) 33𝑥 − 11𝑥 2 = 0
𝑐) 27𝑐 2 + 63𝑐 = 0
𝑑) 100𝑝 + 40𝑝2 = 0
3. The height of a flare above the ground is a function of the time it is in the air. The height
h, in metres, after t seconds, is ℎ = −5𝑡 2 + 100𝑡.
a) Use common factoring to determine the times at which the flare is on the ground.
b) Determine the coordinates of the vertex. What is the maximum height of the flare
and the time it took to reach that maximum height?
Assignment 4: Answers
1. 𝑎) 6(𝑎 + 3)
𝑒) 2(𝑑2 + 2𝑑 + 1)
𝑏) 3𝑥(𝑥 − 3)
𝑓) 8(2𝑥 2 − 4𝑥 − 1)
𝑐) 9𝑐(4 − 5𝑐)
𝑔) 3(3𝑥 2 + 2𝑥 − 4)
𝑑) 5𝑝(3𝑝 − 11)
ℎ) 8(3𝑑2 + 5𝑑 + 7)
2. 𝑎) 𝑎 = 0 and 𝑎 = 3
𝑏) 𝑥 = 0 and 𝑥 = 3
3. 𝑎) 𝑡 = 0 and 𝑡 = 20 seconds
7
3
5
𝑑) 𝑝 = 0 and 𝑝 = −
2
𝑐) 𝑐 = 0 and c = −
𝑏) (10, 500)
Math 10 Academic: MPM2D
Quadratic Functions: Multiplying Two Binomials
Unit
3: Quadratic Functions
Lesson
5: Multiplying Two Binomials
Expected Duration
Expectations
 find the product of two binomials.
Introduction
In this lesson, we will continue to develop the tools necessary for solving quadratic equations
algebraically. In Lesson 4, we used common factoring to solve quadratic equations. In the next
lesson, we will look at how to factor simple trinomials of the form 𝑥 2 + 𝑏𝑥 + 𝑐. In order to
prepare ourselves for factoring expressions of this type however, we need to examine the
reverse process, which is multiplying two binomials (𝑥 + 𝑟) and (𝑥 + 𝑠), where r and s are
integers.
Multiplying Two Binomials
A polynomial with one term, such as 2𝑥 2 , is called a monomial, and a polynomial with two
terms, such as 3𝑥 − 𝑦, is called a binomial. In grade 9, you worked with products of monomials
and binomials using the distributive law.
Example 1
Find the product of 𝑥(𝑥 + 2) using the distributive law.
Solution
𝑥(𝑥 + 2) = 𝑥(𝑥) + 2𝑥
= 𝑥 2 + 2𝑥
When we multiply two binomials, we also use the distributive law. The only difference is that
we use it twice.
Example 2
Find the product of (𝑥 + 3)(𝑥 + 2).
Solution
Multiply each term in the first factor (𝒙 + 𝟑) by the second factor (𝑥 + 2).
(𝒙 + 𝟑)(𝑥 + 2) = 𝒙(𝑥 + 2) + 𝟑(𝑥 + 2)
𝑵𝒐𝒘 𝒖𝒔𝒆 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒗𝒆 𝒍𝒂𝒘 𝒐𝒏 𝒕𝒉𝒆 𝒙(𝒙 + 𝟐) 𝒂𝒏𝒅 𝒕𝒉𝒆 𝟑(𝒙 + 𝟐).
(𝑥 + 3)(𝑥 + 2) = 𝑥 2 + 2𝑥 + 3𝑥 + 6
𝑹𝒆𝒄𝒂𝒍𝒍 (𝒙)(𝒙) = (𝒙𝟏 (𝒙𝟏 ) = 𝒙𝟏+𝟏 = 𝒙𝟐
𝑨𝒅𝒅 𝒕𝒉𝒆 𝒍𝒊𝒌𝒆 𝒕𝒆𝒓𝒎𝒔: 𝟐𝒙 + 𝟑𝒙 = 𝟓𝒙
(𝑥 + 3)(𝑥 + 2) = 𝑥 2 + 5𝑥 + 6
In the second line, we see that the product of two binomials has four terms. This is because
the number of terms in the product of two polynomials equals the number of terms in the first
factor multiplied by the number of terms in the second factor. In this question, each factor is
a binomials so has 2 terms. Therefore the product has 2 × 2 = 4 terms. Since two of those
four terms are like terms, we can combine them so that the final answer is a trinomial.
We can also illustrate the product of (𝑥 + 3)(𝑥 + 2) with a 2×2 table (2 rows × 2 columns).
The entries are the four terms in the product.
𝒙
𝟑
𝒙
𝑥2
3𝑥
𝟐
2𝑥
6
The table shows that (𝑥 + 3)(𝑥 + 2) = 𝑥 2 + 2𝑥 + 3𝑥 + 6
= 𝑥 2 + 5𝑥 + 6
Example 3
Expand (3𝑥 + 1)(2𝑥 + 5).
Solution
(𝟑𝒙 + 𝟏)(2𝑥 + 5) = 𝟑𝒙(2𝑥 + 5) + 𝟏(2𝑥 + 5)
= 3𝑥(2𝑥) + 3𝑥(5) + 1(2𝑥) + 1(5)
= 6𝑥 2 + 15𝑥 + 2𝑥 + 5
= 6𝑥 2 + 17𝑥 + 5
Example 4
Expand (4𝑥 − 5)(3𝑥 + 2).
Solution
(𝟒𝒙 − 𝟓)(3𝑥 + 2) = 𝟒𝒙(3𝑥 + 2) − 𝟓(3𝑥 + 2)
= 4𝑥(3𝑥) + 4𝑥(2) − 5(3𝑥) − 5(2)
= 12𝑥 2 + 8𝑥 − 15𝑥 − 10
= 12𝑥 2 − 7𝑥 − 10
Example 5
Expand (𝑥 − 5)2 .
Solution
(𝑥 + 5)2 = (𝒙 + 𝟓)(𝑥 + 5)
= 𝒙(𝑥 + 5) + 𝟓(𝑥 + 5)
= 𝑥 2 + 5𝑥 + 5𝑥 + 25
(𝑥 + 5)2 = 𝑥 2 + 10𝑥 + 25
In Example 5, (𝑥 + 5)2 is really the product of two identical binomials factors (𝑥 +
5) and (𝑥 + 5). The product of two identical binomials is called a perfect square trinomial.
In general,
(𝒙 + 𝒚)𝟐 = 𝒙𝟐 + 𝟐𝒙𝒚 + 𝒚𝟐 𝐨𝐫 (𝒙 − 𝒚)𝟐 = 𝒙𝟐 − 𝟐𝒙𝒚 + 𝒚𝟐
When we multiply two identical binomials, we apply this formula in one step. Let's do an
example.
Example 6
Expand each binomial square.
𝑎) (𝑥 + 4)2
𝑏) (2𝑥 − 3)2
Solution
𝑎) (𝑥 + 4)2 = 𝑥 2 + 2(𝑥)(4) + 42
𝑏) (2𝑥 − 3)2 = (2𝑥)2 − 2(2𝑥)(3) + 32
Assignment 5: Multiplying Two Binomials
1. Expand each of the following.
𝑎) (𝑥 + 1)(𝑥 + 2)
𝑏) (𝑥 − 4)(𝑥 − 1)
𝑐) (2𝑥 + 1)(𝑥 + 4)
𝑑) (3𝑥 − 2)(𝑥 − 3)
𝑒) (6𝑥 − 7)(3𝑥 − 1)
𝑓) (5𝑥 + 3)(4𝑥 − 5)
𝑔) (3𝑥 + 1)(3𝑥 − 1)
ℎ) (4𝑥 − 9)(4𝑥 + 9)
𝑖) (𝑥 + 1)2
𝑗) (4𝑥 − 5)2
𝑘) (𝑎𝑥 + 𝑏)2
𝑙) (𝑎𝑥 + 𝑏)(𝑎𝑥 − 𝑏)
2. The length and width of a rectangle are 3𝑥 − 4 and 2𝑥 + 1.
a) Find an expression for the area in the form 𝐴 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐.
b) Calculate the area when 𝑥 = 10.
Assignment 5: Answers
1. 𝑎) 𝑥 2 + 3𝑥 + 2
𝑏) 𝑥 2 − 5𝑥 + 4
𝑐) 2𝑥 2 + 9𝑥 + 4
𝑑) 3𝑥 2 − 11𝑥 + 6
𝑒) 18𝑥 2 − 27𝑥 + 7
𝑓) 20𝑥 2 − 13𝑥 − 15
𝑔) 9𝑥 2 − 1
ℎ) 16𝑥 2 − 81
𝑖) 𝑥 2 + 2𝑥 + 1
𝑗) 16𝑥 2 − 40𝑥 + 25
𝑘) 𝑎2 𝑥 2 + 2𝑎𝑏𝑥 + 𝑏 2
𝑙) 𝑎2 𝑥 2 − 𝑏 2
2. 𝑎) 𝐴 = 6𝑥 2 − 5𝑥 − 4
𝑏) 𝐴 = 546
Math 10 Academic: MPM2D
Quadratic Functions: Factoring Simple Trinomials
Unit
3: Quadratic Functions
Lesson
6: Factoring Trinomials of the Form 𝒙𝟐 + 𝒃𝒙 + 𝒄
Expected Duration
Expectations
 to be able to factor simple trinomials in order to solve quadratic equations.
Introduction
In Lesson 5, we learned that multiplying two binomials (𝑥 + 𝑟) and (𝑥 + 𝑠) results in a
quadratic trinomial. In this lesson, we will learn the reverse process - factoring. A simple
quadratic trinomial is a trinomial in the form 𝑥 2 + 𝑏𝑥 + 𝑐. In order to factor such a trinomial,
we need to write it as a product of two binomials. Factoring simple trinomials is one of the
tools we need to algebraically solve quadratic equations.
Factoring Simple Trinomials
To factor a simple trinomial such as 𝑥 2 + 5𝑥 + 6, we need to write it as the product of the two
binomials (𝑥 + 2)(𝑥 + 3). Let's examine how the numbers 2 and 3 in the factors
(𝑥 + 2) and (𝑥 + 3) are related to the numbers 5 and 6 in the trinomial 𝑥 2 + 5𝑥 + 6.
Let's expand (𝑥 + 2)(𝑥 + 3) as we did in Lesson 5.
(𝑥 + 2)(𝑥 + 3) = 𝑥(𝑥 + 3) + 2(𝑥 + 3)
= 𝑥(𝑥 + 3) + 𝟐(𝑥 + 𝟑)
= 𝑥 2 + 𝟑𝒙 + 𝟐𝒙 + 𝟔
= 𝑥 2 + 𝟓𝒙 + 𝟔
From the expansion, we see that the coefficient 𝟓 of the middle term 𝟓𝒙 in the trinomial is the
sum of the numbers 𝟐 and 𝟑 of the factors (𝑥 + 2) and (𝑥 + 3).
The third term 𝟔 in the trinomial is the product of the numbers 𝟐 and 𝟑.
Therefore, in order to factor a simple trinomial such as 𝑥 2 + 5𝑥 + 6, we need to find factors
of the last term 6 that have a sum of 5. Those factors are 2 and 3, so the factored form is
(𝑥 + 2)(𝑥 + 3).
In general:
In order to factor 𝒙𝟐 + 𝒃𝒙 + 𝒄, we need to find factors of c that have a sum of b.
Let those factors of c be r and s. Then 𝒙𝟐 + 𝒃𝒙 + 𝒄 = (𝒙 + 𝒓)(𝒙 + 𝒔).
Example 1
Factor 𝑥 2 + 9𝑥 + 20.
Solution
We need to find factors of 20 that have a sum of 9. Since the product (20) is a positive number,
the factors have the same sign. Since the sum (9) is also positive, both factors must be positive.
Let's make a list of pairs factors of 20 until we find the pair that has a sum of 9.
1 × 20
This pair has a sum of 1 + 20 = 21, so it is not the correct pair.
2 × 10
This pair has a sum of 2 + 10 = 12, so it is not the correct pair.
4×5
This pair has a sum of 4 + 5 = 9, so it is the correct pair.
Therefore, 𝑥 2 + 9𝑥 + 20 = (𝑥 + 4)(𝑥 + 5)
Example 2
Factor 𝑥 2 − 3𝑥 − 28
Solution
We need to find factors of -28 that have a sum of -3. Since the product (-28) is a negative
number, the factors have opposite signs: One factor is negative and the other factor is positive.
Let's make a list of pairs factors of -28 until we find the pair that has a sum of -3. We'll mark
incorrect pairs with an ×.
−1 × 28
Sum = −1 + 28 = 27 ×
1 × (−28)
Sum = 1 + (−28) = 1 − 28 = −27 ×
−2 × 14
Sum = −2 + 14 = 12 ×
2 × (−14)
Sum = 2 + (−14) = 2 − 14 = −12 ×
−4 × 7
Sum = −4 + 7 = 3 ×
4 × (−7)
Sum = 4 + (−7) = 4 − 7 = −3
This is correct!
Therefore, 𝑥 2 − 3𝑥 − 28 = (𝑥 + 4)(𝑥 − 7)
Example 3
Factor 𝑥 2 − 20𝑥 + 75
Solution
We need to find factors of 75 that have a sum of -20. Since the product (75) is positive, the
factors have the same sign. Since the sum (-20) is negative, both factors must be negative. Let's
make a list of pairs of negative factors of 75 until we find the pair that has a sum of -20. We'll
mark incorrect pairs with an ×.
−1 × (−75)
Sum = −1 + (−75) = −1 − 75 = −76 ×
−3 × (−25)
Sum = −3 + (−25) = −3 − 25 = −28 ×
−5 × (−15)
Sum = −5 + (−15) = −5 − 15 = −20 Correct
Therefore, 𝑥 2 − 20𝑥 + 75 = (𝑥 − 5)(𝑥 − 15)
Example 4
Factor 𝑥 2 + 10𝑥 + 25.
Solution
Recall from Lesson 5: The product of two identical binomials is called a perfect square
trinomial, where (𝒙 + 𝒚)𝟐 = 𝒙𝟐 + 𝟐𝒙𝒚 + 𝒚𝟐 𝐨𝐫 (𝒙 − 𝒚)𝟐 = 𝒙𝟐 − 𝟐𝒙𝒚 + 𝒚𝟐 .
When we are factoring quadratic trinomials, we need to look out for perfect square trinomials.
These are trinomials that have the form 𝑥 2 + 2𝑥𝑦 + 𝑦 2 or 𝑥 2 − 2𝑥𝑦 + 𝑦 2 . If we can identify
them, we can factor them in one step. The trinomial 𝑥 2 + 10𝑥 + 25 is a perfect square
trinomial. Let's show this by comparing 𝑥 2 + 10𝑥 + 25 to 𝑥 2 + 2𝑥𝑦 + 𝑦 2 . First of all, we see
that both the first (𝑥 2 ) and third (25) terms in 𝑥 2 + 10𝑥 + 25 are perfect squares as are the
first and third terms in 𝑥 2 + 2𝑥𝑦 + 𝑦 2 . If we equate the third terms 𝑦 2 = 25, we see that 𝑦 =
5 and the coefficient 10 of the second term in 𝑥 2 + 10𝑥 + 25 is 2 × 5 = 2𝑦 as is the
coefficient 2𝑦 of the second term in 𝑥 2 + 2𝑥𝑦 + 𝑦 2 .
Therefore, 𝑥 2 + 10𝑥 + 25 is a perfect square trinomial. The number we use in the brackets is
the square root of 25, which is 5.
𝑥 2 + 10𝑥 + 25 = (𝑥 + 5)2
Example 5
Factor 𝑥 2 − 14𝑥 + 49.
Solution
This is a perfect square trinomial of the form 𝒙𝟐 − 𝟐𝒙𝒚 + 𝒚𝟐 , since 49 is a perfect square
whose square root is 7 and the coefficient of the second term is 2 × 7 = 14. Therefore,
𝑥 2 − 14𝑥 + 49 = (𝑥 − 7)2
Example 6
Factor 2𝑥 2 − 26𝑥 + 60.
Solution
Often more than one type of factoring is required in order to fully factor a trinomial. For any
question that involves factoring, it is important to always check for a common factor first. Then
factor out the GCF before using any other factoring method. The trinomial
2𝑥 2 −
26𝑥 + 60 has a GCF of 2, so we factor 2 out of every term before doing anything else.
2𝑥 2 26𝑥 60
2
2𝑥 − 26𝑥 + 60 = 2 (
−
+ )
2
2
2
= 2(𝑥 2 − 13𝑥 + 30)
Now we have a simple trinomial in the brackets. We need to find factors of 30 that have a sum
of -13. Since the product (30) is positive, the factors have the same sign. (They are either both
positive or both negative.) Since the sum (-13) is negative, both factors must be negative. Let's
make a list of pairs of negative factors of 30 until we find the pair that has a sum of -13. We'll
mark incorrect pairs with an ×.
−1 × (−30)
Sum = −1 + (−30) = −1 − 30 = −31 ×
−2 × (−15)
Sum = −2 + (−15) = −2 − 15 = −17 ×
−3 × (−10)
Sum = −3 + (−10) = −3 − 10 = −13 Correct
Therefore, 2𝑥 2 − 26𝑥 + 60 = 2(𝑥 2 − 13𝑥 + 30)
= 2(𝑥 − 3)(𝑥 − 10)
Example 7
Factor 𝑥 3 + 11𝑥 2 − 26𝑥.
Solution
This trinomial has a GCF of 𝑥, so we factor 𝑥 out of every term before doing anything else.
𝑥 3 11𝑥 2 26𝑥
𝑥 + 11𝑥 − 26𝑥 = 𝑥 ( +
−
)
𝑥
𝑥
𝑥
3
2
= 𝑥(𝑥 2 + 11𝑥 − 26)
Now we have a simple trinomial in the brackets. We need to find factors of -26 that have a
sum of 11. Since the product (-26) is negative, the factors have opposite signs.
−1 × 26
Sum = −1 + 26 = 25 ×
1 × (−26)
Sum = 1 + (−26) = 1 − 26 = −25 ×
−2 × 13
Sum = −2 + 13 = 11 Correct
Therefore, 𝑥 3 + 11𝑥 2 − 26𝑥 = 𝑥(𝑥 2 + 11𝑥 − 26)
= 𝑥(𝑥 − 2)(𝑥 + 13)
Example 8
Solve the equation 𝑥 2 + 7𝑥 + 10 = 0.
Solution
𝑥 2 + 7𝑥 + 10 = 0
To factor the left side, we need to find factors of 10 that have a sum of 7. Since the product
and sum are both positive, the factors are both positive.
1 × 10
Sum = 1 + 10 = 11 ×
2×5
Sum = 2 + 5 = 5 Correct
Therefore the left side factors as (𝑥 + 2)(𝑥 + 5) and we can find the roots of the equation.
𝑥 2 + 7𝑥 + 10 = 0
(𝑥 + 2)(𝑥 + 5) = 0
To solve for the roots, we set both factors equal to zero.
Therefore 𝑥 + 2 = 0
and
𝑥 = −2 and
𝑥+5=0
𝑥 = −5
The roots of the equation are 𝑥 = −2 and 𝑥 = −5.
Example 9
Solve the equation 3𝑥 2 − 36𝑥 + 108 = 0.
Solution
3𝑥 2 − 36𝑥 + 108 = 0
Since the left side has a GCF of 3, we need to factor 3 out of each term on the left side.
3𝑥 2 − 36𝑥 + 108 = 0
3𝑥 2 36𝑥 108
3(
−
+
)=0
3
3
3
3(𝑥 2 − 12𝑥 + 36) = 0
We can divide both sides of the equation by 3.
3(𝑥 2 − 12𝑥 + 36) 0
=
3
3
𝑥 2 − 12𝑥 + 36 = 0
The left side is a perfect square trinomial of the form 𝒙𝟐 − 𝟐𝒙𝒚 + 𝒚𝟐 , since 36 is a perfect
square whose square root is 6 and the coefficient of the second term is 2 × 6 = 12. Therefore,
𝑥 2 − 12𝑥 + 36 = 0
becomes
(𝑥 − 6)2 = 0
This equation has only one root. We can find that root by setting 𝑥 − 6 = 0.
𝑥−6=0
𝑥=6
The root of the equation is 𝑥 = 6.
Example 10
Solve the equation 2𝑥 2 − 8𝑥 − 24 = 0.
Solution
2𝑥 2 − 8𝑥 − 24 = 0
First we factor out the GCF of 2.
2𝑥 2 8𝑥 24
2(
−
− )=0
2
2
2
2(𝑥 2 − 4𝑥 − 12) = 0
We can divide both sides of the equation by 2.
2(𝑥 2 − 4𝑥 − 12) 0
=
2
2
𝑥 2 − 4𝑥 − 12 = 0
We need to find factors of -12 that have a sum of -4. Since the product (-12) is negative, the
factors have opposite signs. Let's make a list of pairs factors of -12 until we find the pair that
has a sum of -4.
−1 × 12
Sum = −1 + 12 = 11 ×
1 × (−12)
Sum = 1 + (−12) = 1 − 12 = −11 ×
−2 × 6
Sum = −2 + 6 = 4
2 × (−6)
Sum = 2 + (−6) = 2 − 6 = −4 Correct
×
Therefore, the left side factors as (𝑥 + 2)(𝑥 − 6) and we can find the roots.
𝑥 2 − 4𝑥 − 12 = 0
becomes
(𝑥 + 2)(𝑥 − 6) = 0
To solve for the roots, we set both factors equal to zero.
𝑥+2=0
and
𝑥−6=0
𝑥 = −2 and
𝑥=6
Assignment 6: Factoring Simple Trinomials
Assignment 6: Answers
Math 10 Academic: MPM2D
Quadratic Functions: Factoring Trinomials of the Form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄
Unit
3: Quadratic Functions
Lesson
7: Factoring Trinomials of the Form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄
Expected Duration
Expectations
 to be able to factor complex trinomials in order to solve quadratic equations.
Introduction
In Lesson 6, we factored simple quadratic trinomials of the form 𝒙𝟐 + 𝒃𝒙 + 𝒄, in which the
coefficient of the squared term is 1. When the coefficient of the squared term is not 1, the
trinomial is in the form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 and is called a complex quadratic trinomial. There are
two methods we can use to factor a complex trinomial: Factoring by Decomposition and the
Australian Method. Factoring a simple trinomial and the two methods we will use to factor
complex trinomials all share a common feature. We always need to find the product and sum
of the integers that allow the trinomial to be written in factored form. When we factor a
complex trinomial, we write it as the product of two binomials (𝑝𝑥 + 𝑟) and (𝑞𝑥 + 𝑠).
Factoring Complex Trinomials
A. Factoring by Decomposition
When we factor a simple trinomial, 𝑥 2 + 𝑏𝑥 + 𝑐, we need to find factors of c that have a sum
of b. The product of the factors is equal to c and the sum of the factors is equal to 𝑏.
𝑻𝒐 𝒇𝒂𝒄𝒕𝒐𝒓 𝒙𝟐 + 𝒃𝒙 + 𝒄,
𝒑𝒓𝒐𝒅𝒖𝒄𝒕 = 𝒄 𝒂𝒏𝒅 𝒔𝒖𝒎 = 𝒃
When we factor a complex trinomial, 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, we also need to find factors of the product
that have a sum of b. The difference is that for a complex trinomial, the product of the factors
is equal to 𝑎 × 𝑐, rather than just 𝑐. When we find the factors of 𝑎 × 𝑐 that have a sum of 𝑏,
we need to break down, or decompose, the second term 𝑏𝑥 into those factors. Let's do an
example.
Example 1
Factor 3𝑥 2 + 17𝑥 + 10.
Solution
In the trinomial, 3𝑥 2 + 17𝑥 + 10, the values of 𝑎, 𝑏, and 𝑐 are 𝑎 = 3, 𝑏 = 17, and 𝑐 = 10.
Product of the factors = 𝑎 × 𝑐 = 3 × 10 = 30
Sum of the factors = 𝑏 = 17
Since both the product and sum are positive, both factors are positive. We need to make a list
of pairs of factors of 30 until we find the pair that has a sum of 17.
1 × 30
Sum = 1 + 30 = 31 ×
2 × 15
Sum = 2 + 15 = 17 Correct
Now we break down the second term (17𝑥) of the trinomial 3𝑥 2 + 17𝑥 + 10 into 𝑥 terms
whose coefficients are the two factors we have found. So we replace 17𝑥 with 2𝑥 + 15𝑥.
3𝑥 2 + 17𝑥 + 10 = 3𝑥 2 + 2𝑥 + 15𝑥 + 10
Now we factor out the GCF from the first two terms 3𝑥 2 + 2𝑥 and the GCF from the last two
terms 15𝑥 + 10. The GCF for the first two terms is 𝑥 and the GCF for the last two terms is 5.
3𝑥 2 + 17𝑥 + 10 = 3𝑥 2 + 2𝑥 + 15𝑥 + 10
3𝑥 2 2𝑥
15𝑥 10
= 𝑥(
+ )+ 5(
+ )
𝑥
𝑥
5
5
= 𝑥(3𝑥 + 2) + 5(3𝑥 + 2)
There are two terms in the expression above; 𝑥(3𝑥 + 2) and 5(3𝑥 + 2). The GCF for these
terms is a binomial (3𝑥 + 2). The presence of a binomial common factor at this stage of the
solution indicates that our solution so far is correct. Now we factor out the common factor
(3𝑥 + 2) from each term.
3𝑥 2 + 17𝑥 + 10 = 𝒙(3𝑥 + 2) + 𝟓(3𝑥 + 2)
= (3𝑥 + 2)(𝒙 + 𝟓)
Therefore, 3𝑥 2 + 17𝑥 + 10 = (3𝑥 + 2)(𝑥 + 5)
Let's summarize the steps that are involved in factoring a complex trinomial by decomposition.
Factoring 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 by Decomposition
1. Identify the product and sum of the factors: 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 = 𝑎 × 𝑐 and 𝑆𝑢𝑚 = 𝑏
2. Find the factors with the product and sum identified in Step 1.
3. Decompose the 𝑥 term in the trinomial into two terms whose coefficients are the factors
found in Step 2. The expression will now have four terms.
4. Factor out the GCF from the first two terms and the GCF from the last two terms.
5. Factor out the binomial common factor from each term.
Example 2
Factor 7𝑎2 − 29𝑎 + 24.
Solution
In 7𝑎2 − 29𝑎 + 24, the values of 𝑎, 𝑏, and 𝑐 are 𝑎 = 7, 𝑏 = −29, and 𝑐 = 24.
Product of the factors = 𝑎 × 𝑐 = 7 × 24 = 168
Sum of the factors = 𝑏 = −29
Since the product (168) is positive, the factors have the same sign. Since the sum (-29) is
negative, the factors must be negative. Let's make a list of possible factor pairs.
−1 × (−168)
Sum = −1 + (−168) = −1 − 168 = −169 ×
−2 × (−84)
Sum = −2 + (−84) = −2 − 84 = −86
×
−3 × (−56)
Sum = −3 + (−56) = −3 − 56 = −59
×
−4 × (−42)
Sum = −4 + (−42) = −4 − 42 = −46
×
−6 × (−28)
Sum = −6 + (−28) = −6 − 28 = −34
×
−7 × (−24)
Sum = −7 + (−24) = −7 − 24 = −31
×
−8 × (−21)
Sum = −8 + (−21) = −8 − 21 = −29 Correct!
Now we decompose the second term (−29𝑎) of the trinomial into two 𝑎 terms whose
coefficients are the factors −8 and − 21. So we replace −29𝑎 with −8𝑎 − 21𝑎.
7𝑎2 − 29𝑎 + 24 = 7𝑎2 − 8𝑎 − 21𝑎 + 24
We factor out the GCF from the first two terms 7𝑎2 − 8𝑎 and the GCF from the last two terms
−21𝑎 + 24. The GCF for the first two terms is 𝑎 and for the last two terms is −3.
7𝑎2 − 29𝑎 + 24 = 7𝑎2 − 8𝑎 − 21𝑎 + 24
= 𝑎(7𝑎 − 8) − 3(7𝑎 − 8)
Now we factor out the common factor (7𝑎 − 8) from each term.
7𝑎2 − 29𝑎 + 24 = 𝑎(7𝑎 − 8) − 3(7𝑎 − 8)
= (7𝑎 − 8)(𝑎 − 3)
Example 3
Factor 9𝑤 2 + 12𝑤 + 4.
Solution
9𝑤 2 + 12𝑤 + 4 is a perfect square trinomial of the form 𝑥 2 + 2𝑥𝑦 + 𝑦 2 , since 9 = 𝟑2 , 4 =
𝟐2 , and the coefficient (12) of the second term is equal to 2 × 𝟑 × 𝟐. We can factor this
trinomial in the form (𝑥 + 𝑦)2 , where 𝑥 is 𝟑𝑤 and 𝑦 is 𝟐.
9𝑤 2 + 12𝑤 + 4 = (3𝑤 + 2)2
This trinomial can also be factored by decomposition, but recognizing it as a perfect square
trinomial allows for a more direct solution.
Example 4
Factor 14𝑥 2 − 62𝑥 − 40.
Solution
The trinomial 14𝑥 2 − 62𝑥 − 40 has a GCF of 2, so we need to factor 2 out of each term.
14𝑥 2 − 62𝑥 − 40 = 2(7𝑥 2 − 31𝑥 − 20)
In the trinomial in the brackets, 𝑎 = 7, 𝑏 = −31, and 𝑐 = −20.
Product of the factors = 𝑎 × 𝑐 = 7 × (−20) = −140
Sum of the factors = 𝑏 = −31
Since the product (−140) is negative, the factors have opposite signs. Since the sum (−31) is
negative, it will be the largest factor that is negative. Keeping this in mind will reduce the
number of possible factor pairs.
1 × (−140)
Sum = 1 + (−140) = 1 − 140 = −139
2 × (−70)
Sum = 2 + (−70) = 2 − 70 = −68
4 × (−35)
Sum = 4 + (−35) = 4 − 35 = −31 Correct
×
×
Now we decompose the second term (−31𝑥) of the trinomial 7𝑥 2 − 31𝑥 − 20 into two 𝑥
terms whose coefficients are the factors 4 and − 35. So we replace −31𝑥 with 4𝑥 − 35𝑥.
14𝑥 2 − 62𝑥 − 40 = 2(7𝑥 2 − 31𝑥 − 20)
= 2(7𝑥 2 + 4𝑥 − 35𝑥 − 20)
We factor out the GCF from the first two terms 7𝑥 2 + 4𝑥 and the GCF from the last two terms
−35𝑥 − 20. The GCF for the first two terms is 𝑥 and for the last two terms is −5.
14𝑥 2 − 62𝑥 − 40 = 2[𝑥(7𝑥 + 4) − 5(7𝑥 + 4)]
= 2[(7𝑥 + 4)(𝑥 − 5)]
= 2(7𝑥 + 4)(𝑥 − 5)
Example 5
Solve 11𝑝2 − 95𝑝 − 36 = 0.
Solution
11𝑝2 − 95𝑝 − 36 = 0
Since the first coefficient 11 is a prime number, the left side does not have a common factor,
so we find the product and sum of the factors and factor by decomposition.
Since 𝑎 = 11, 𝑏 = −95, and 𝑐 = −36,
Product of the factors = 𝑎 × 𝑐 = 11 × (−36) = −396
Sum of the factors = 𝑏 = −95
Since the product (−396) is negative, the factors have opposite signs. Since the sum (−95) is
negative, it will be the largest factor that is negative.
1 × (−396)
Sum = 1 + (−396) = 1 − 396 = −395
×
2 × (−198)
Sum = 2 + (−198) = 2 − 198 = −196
×
3 × (−132)
Sum = 3 + (−132) = 3 − 132 = −129
×
4 × (−99)
Sum = 4 + (−99) = 4 − 99 = −95 Correct
The factors are 4 and − 99. We decompose the second term (−95𝑝) of the left side of the
equation into two 𝑝 terms whose coefficients are the factors 4 and − 99. So we replace −95𝑝
with 4𝑝 − 99𝑝.
11𝑝2 − 95𝑝 − 36 = 0
11𝑝2 + 4𝑝 − 99𝑝 − 36 = 0
𝑝(11𝑝 + 4) − 9(11𝑝 + 4) = 0
(11𝑝 + 4)(𝑝 − 9) = 0
Therefore,
11𝑝 + 4 = 0
and
𝑝−9=0
11𝑝 = −4
𝑝=9
11𝑝 −4
=
11
11
4
𝑝=−
11
The roots of the equation are 𝑝 = −
4
11
and p = 9.
A. The Australian Method of Factoring
The Australian method and the method of decomposition share many common features. Both
are used to factor complex trinomials. Both require that we find factors of the product 𝑎 × 𝑐
that have a sum of b. The point at which the methods diverge is what we do after having found
the factors. When we factor by decomposition, we decompose the second term of the
trinomial into two terms whose coefficients are the factors. When we use the Australian
Method however, we substitute the factors 𝑟 and 𝑠 into the following expression.
(𝑎𝑥 + 𝑟)(𝑎𝑥 + 𝑠)
𝑎
Then we factor out the GCF from one or both of the factors in the numerator and simplify. The
final answer is identical to what we would obtain using the decomposition method. Let's do
some examples together.
Example 6
Use the Australian method to factor 7𝑎2 − 29𝑎 + 24.
Solution
We previously factored this trinomial using the method of decomposition in Example 2. We
are going to show that we can obtain the same solution using the Australian method. In
Example 2, we calculated the product and sum of the factors as follows.
Product of the factors = 𝑎 × 𝑐 = 7 × 24 = 168
Sum of the factors = 𝑏 = −29
We found the correct factors to be −8 and − 21, since Product = (−8)(−21) = 168 and
Sum = −8 + (−21) = −8 − 21 = −29.
Now we substitute 𝑎 = 7 and the factors 𝑟 = −8 and s = −21 into the following expression.
(𝑎𝑥 + 𝑟)(𝑎𝑥 + 𝑠)
𝑎
(7𝑥 + (−8))(7𝑥 + (−21))
=
7
(7𝑥 − 8)(7𝑥 − 21)
=
7
7𝑎2 − 29𝑎 + 24 =
We can factor out the GCF of 7 from the binomial factor (7𝑥 − 21) in the numerator.
(7𝑥 − 8)(7)(𝑥 − 3)
7
7(7𝑥 − 8)(𝑥 − 3)
=
7
7𝑎2 − 29𝑎 + 24 =
Finally we write the fraction in lowest terms. (The 7 in the numerator cancels the 7 in the
denominator.)
7𝑎2 − 29𝑎 + 24 = (7𝑥 − 8)(𝑥 − 3)
The final answer is identical to that obtained using decomposition in Example 2.
Example 7
Use the Australian method to factor 12𝑥 2 − 34𝑥 + 10.
Solution
This trinomial has a GCF of 2. We need to factor out 2 from each term.
12𝑥 2 − 34𝑥 + 10 = 2(6𝑥 2 − 17𝑥 + 5)
Now we factor the trinomial 6𝑥 2 − 17𝑥 + 5 using the Australian method.
Product of the factors = 𝑎 × 𝑐 = 6 × 5 = 30
Sum of the factors = 𝑏 = −17
We need to find factors of 30 that have a sum of -17. Since the product is positive and the sum
is negative, both factors are negative. We'll make a list of factor pairs.
−1 × (−30)
Sum = −1 + (−30) = −1 − 30 = −31
−2 × (−15)
Sum = −2 + (−15) = −2 − 15 = −17 Correct
×
The factors are -2 and -15. We substitute 𝑎 = 6 and the factors 𝑟 = −2 and s = −15 into the
appropriate expression.
12𝑥 2 − 34𝑥 + 10 = 2(6𝑥 2 − 17𝑥 + 5)
= 2[
(𝑎𝑥 + 𝑟)(𝑎𝑥 + 𝑠)
]
𝑎
= 2[
(6𝑥 − 2)(6𝑥 − 15)
]
6
We can factor out the GCF of 2 from the binomial factor (6𝑥 − 2) and the GCF of 3 from
(6𝑥 − 15).
12𝑥 2 − 34𝑥 + 10 = 2 [
= 2[
(2)(3𝑥 − 1)(3)(2𝑥 − 5)
]
6
(6)(3𝑥 − 1)(2𝑥 − 5)
]
6
= 2(3𝑥 − 1)(2𝑥 − 5)
Assignment 7: Factoring Complex Trinomials
Assignment 7: Answers
Math 10 Academic: MPM2D
Quadratic Functions: Factoring a Difference of Squares
Unit
3: Quadratic Functions
Lesson
8: Factoring a Difference of Squares
Expected Duration
Expectations

to be able to factor a difference of squares.
Introduction
In Lesson 6, we factored simple trinomials in Lesson 7, we factored complex trinomials of the
form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 in which the coefficient of the squared term is not 1. The last type of
factoring we will discuss is referred to as difference of squares. This is a binomial that can be
expressed in the form 𝑥 2 − 𝑦 2 .
Factoring a Difference of Squares
A difference of squares such as 𝑥 2 − 𝑦 2 can be factored as the product of the sum and
difference of 𝑥 and 𝑦:
𝒙𝟐 − 𝒚𝟐 = (𝒙 + 𝒚)(𝒙 − 𝒚)
A difference of squares consists of two terms, both of which are perfect squares, and the terms
are separated by a subtraction sign. Let's do some examples.
Example 1
Factor each of the following.
𝑎) 𝑥 2 − 4
𝑏) 5𝑥 2 − 45
Solution
𝑎) 𝑥 2 − 4 = (𝑥 + 2)(𝑥 − 2)
𝑏) Factor out the common factor of 5.
5𝑥 2 − 45 = 5(𝑥 2 − 9)
= 5(𝑥 + 3)(𝑥 − 3)
Example 2
Factor each of the following.
𝑎) 32𝑥 2 − 50𝑦 2
𝑏) 4𝑎3 − 196𝑎𝑏 2
Solution
Factor out the GCF first.
𝑎) 32𝑥 2 − 50𝑦 2 = 2(16𝑥 2 − 25𝑦 2 )
= 2(4𝑥 + 5𝑦)(4𝑥 − 5𝑦)
𝑏) 4𝑎3 − 196𝑎𝑏 2 = 4𝑎(𝑎2 − 49𝑏 2 )
= 4𝑎(𝑎 + 7𝑏)(𝑎 − 7𝑏)
Example 3
Factor each of the following.
𝑎) (𝑥 + 2)2 − 𝑦 2
𝑏) 𝑎2 − (3𝑎 + 1)2
Solution
𝑎) (𝒙 + 𝟐)2 − 𝑦 2 = [(𝒙 + 𝟐) + 𝑦 ][(𝒙 + 𝟐) − 𝑦 ]
= (𝑥 + 𝑦 + 2)(𝑥 − 𝑦 + 2)
𝑏) 𝑎2 − (𝟑𝒂 + 𝟏)2 = [𝑎 + (𝟑𝒂 + 𝟏) ][𝑎 − (𝟑𝒂 + 𝟏) ]
= (𝑎 + 3𝑎 + 1)(𝑎 − 3𝑎 − 1)
= (4𝑎 + 1)(−2𝑎 − 1)
Assignment 8: Factoring Difference of Squares
Assignment 8: Answers
Math 10 Academic: MPM2D
Quadratic Functions: Solving Quadratic Equations by Factoring
Unit
3: Quadratic Functions
Lesson
9: Solving Quadratic Equations by Factoring
Expected Duration
Expectations
 to be able to solve a quadratic equation by using factoring methods.
Introduction
This lesson is a culmination of Lessons 4, 6, 7, and 8. We will use the four types of factoring:
common factoring, factoring Simple Trinomials, factoring complex trinomials, and factoring
difference of squares, in order to solve quadratic equations and real-life applications of
quadratic equations.
Solving Quadratic Equations by Factoring
Now that we have learned several methods of factoring, we have developed the tools
necessary to solve quadratic equations algebraically.
Example 1
Solve each equation. Remember to check for a common factor.
𝑎) 𝑥 2 + 9𝑥 + 18 = 0
𝑏) 𝑥 2 − 10𝑥 + 16 = 0
𝑐) 𝑥 2 − 𝑥 − 42 = 0
𝑑) 2𝑥 2 + 6𝑥 − 108 = 0
Solution
𝑎) 𝑥 2 + 9𝑥 + 18 = 0
𝑏) 𝑥 2 − 10𝑥 + 16 = 0
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = 18
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = 16
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = 9
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = −10
The factors are 3 and 6.
The factors are − 2 and − 8.
(𝑥 + 3)(𝑥 + 6) = 0
(𝑥 − 2)(𝑥 − 8) = 0
𝑥 + 3 = 0 and 𝑥 + 6 = 0
𝑥 = −3
𝑥 = −6
The roots are 𝑥 = −3 and 𝑥 = −6.
𝑐) 𝑥 2 − 𝑥 − 42 = 0
𝑥 − 2 = 0 and 𝑥 − 6 = 0
𝑥=2
𝑥=6
The roots are 𝑥 = 2 and 𝑥 = 6.
𝑑) 2𝑥 2 + 6𝑥 − 108 = 0
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = −42
Factor out 2 from each term.
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = −1
2(𝑥 2 + 3𝑥 − 54) = 0
The factors are 6 and − 7.
Divide both sides by 2.
(𝑥 + 6)(𝑥 − 7) = 0
2(𝑥 2 + 3𝑥 − 54) 0
=
2
0
𝑥 + 6 = 0 and 𝑥 − 7 = 0
𝑥 = −6
𝑥=7
The roots are 𝑥 = −6 and 𝑥 = 7.
𝑥 2 + 3𝑥 − 54 = 0
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 = −54
𝑆𝑢𝑚 = 3
The factors are 9 and − 6.
Example 2
Solve each equation. Remember to check for a common factor. If necessary, rewrite the
equation so that all terms are on the left side.
𝑎) 4𝑥 2 − 28𝑥 + 49 = 0
𝑏) 2𝑥 2 = 18𝑥
𝑐) 𝑥 2 = 32 − 4𝑥
𝑑) 27𝑥 2 = −144𝑥 − 192
Solution
𝑎) 4𝑥 2 − 28𝑥 + 49 = 0
𝑏) 2𝑥 2 = 18𝑥
Notice that 2(√4)(√49) = 2(2)(7) = 28
2𝑥 2 − 18𝑥 = 0
Therefore this is a perfect square trinomial.
2𝑥(𝑥 − 9) = 0
Therefore, (2𝑥 − 7)2 = 0
We can dividing both sides by 2.
This equation has 1 root.
𝑥(𝑥 − 9) = 0
2𝑥 − 7 = 0
𝑥 = 0 and 𝑥 − 9 = 0
2𝑥 = 7
𝑥=9
2𝑥 7
=
2
2
7
𝑥=
2
The roots are 𝑥 = 0 and 𝑥 = 9.
7
The root is 𝑥 = .
2
𝑐) 𝑥 2 = 32 − 4𝑥
𝑑) 27𝑥 2 = −144𝑥 − 192
𝑥 2 = 32 − 4𝑥
27𝑥 2 = −144𝑥 − 192
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = 18
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = 9
The factors are 3 and 6.
(𝑥 + 3)(𝑥 + 6) = 0
𝑥 + 3 = 0 and 𝑥 + 6 = 0
𝑥 = −3
𝑥 = −6
The roots are 𝑥 = −3 and 𝑥 = −6.
Assignment 9: Solving Quadratic Equations by Factoring
Assignment 9: Answers
Math 10 Academic: MPM2D
Quadratic Functions: Solving Quadratic Equations by Completing the Square
Unit
Lesson
3: Quadratic Functions
10: Solving Quadratic Equations by Completing the
Square
Expected
Duration
Expectations
 to be able to solve quadratic equations by completing the square.
Introduction
Most quadratic equations cannot be solved by factoring. Many quadratic trinomials do not
have a common factor and cannot be written as the product of two binomials. These trinomials
need to be solved by completing the square. To use this method, the equation must be
rewritten in such a way that the left side is a binomial square: either (𝑥 + 𝑎)2 or (𝑥 − 𝑎)2 .
We'll develop this method through the use of examples.
Solving Quadratic Equations by Completing the Square
Assignment 10: Solving Quadratic Equations by Completing the Square
Assignment 10: Answers
Math 10 Academic: MPM2D
Quadratic Functions: Solving Quadratic Equations by Using a Formula
Unit
3: Quadratic Functions
Lesson
11: Solving Quadratic Equations by Using a Formula
Expected Duration
Expectations
 to be able to solve quadratic equations by using the quadratic formula.
 to explain what the value of the discriminant reveals about the nature of the roots of a
quadratic equation.
Introduction
In Lesson 10, we used the method of completing the square to solve quadratic equations that
could not be solved by factoring. We can apply the method of completing the square to solve
the general quadratic equation 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎. The solution to this equation is a formula
that can be used to solve any quadratic equation that cannot be solved by factoring. By the
end of this lesson, you will be able to use either the method of completing the square or the
quadratic formula to solve quadratic equations that cannot be solved by a factoring method.
We will examine the derivation of the quadratic formula and then we will use it to solve
quadratic equations that cannot be solved by factoring.
We will also examine a part of the formula called the discriminant and what the value of the
discriminant reveals about the nature of the roots of a quadratic equation.
Solving Quadratic Equations Using the Quadratic Formula
Assignment 11: Solving Quadratic Equations Using the Quadratic
Formula
Assignment 11: Answers