# 9-5

```9-5
Solving Quadratic Equations by Factoring
Preview
Warm Up
California Standards
Lesson Presentation
9-5
Solving Quadratic Equations by Factoring
Warm Up
Find each product.
1. (x + 2)(x + 7)
2. (x – 11)(x + 5)
x2 + 9x + 14
x2 – 6x – 55
3. (x – 10)2 x2 – 20x + 100
Factor each polynomial.
4. x2 + 12x + 35
(x + 5)(x + 7)
5. x2 + 2x – 63
(x – 7)(x + 9)
6. x2 – 10x + 16
(x – 2)(x – 8)
7. 2x2 – 16x + 32
2(x – 4)2
9-5
Solving Quadratic Equations by Factoring
California
Standards
14.0 Students solve a quadratic
equation by factoring or completing the
square.
23.0 Students apply quadratic
equations to physical problems, such as
the motion of an object under the force
of gravity.
9-5
Solving Quadratic Equations by Factoring
You have solved quadratic equations by graphing.
Another method used to solve quadratic equations is
to factor and use the Zero Product Property.
9-5
Solving Quadratic Equations by Factoring
Additional Example 1A: Use the Zero Product Property
Use the Zero Product Property to solve the
(x – 7)(x + 2) = 0
x – 7 = 0 or x + 2 = 0
x = 7 or x = –2
Use the Zero Product
Property.
Solve each equation.
9-5
Solving Quadratic Equations by Factoring
Additional Example 1A Continued
Use the Zero Product Property to solve the
Check (x – 7)(x + 2) = 0
(7 – 7)(7 + 2)
(0)(9)
0
0
0
0
Check (x – 7)(x + 2) = 0
(–2 – 7)(–2 + 2)
(–9)(0)
0
0
0
0
Substitute each solution
for x in the original
equation.
9-5
Solving Quadratic Equations by Factoring
Additional Example 1B: Use the Zero Product Property
Use the Zero Product Property to solve each
(x – 2)(x) = 0
(x)(x – 2) = 0
x = 0 or x – 2 = 0
x=2
Check (x – 2)(x) = 0
(0 – 2)(0)
(–2)(0)
0
0
0
0
Use the Zero Product
Property.
Solve the second
equation.
Substitute
(x – 2)(x) =
each solution
(2 – 2)(2)
for x in the
(0)(2)
original
0
equation.
0
0
0
0
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 1a
Use the Zero Product Property to solve each
(x)(x + 4) = 0
x = 0 or x + 4 = 0
x = –4
Check (x)(x + 4) = 0
(0)(0 + 4)
(0)(4)
0
0
0
0
Use the Zero Product
Property.
Solve the second equation.
Substitute
(x)(x +4) = 0
each solution
for x in the (–4)(–4 + 4) 0
(–4)(0) 0
original
equation.
0
0
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 1b
Use the Zero Product Property to solve the
(x + 4)(x – 3) = 0
x + 4 = 0 or x – 3 = 0
x = –4 or
x=3
Use the Zero Product
Property.
Solve each equation.
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 1b Continued
Use the Zero Product Property to solve the
(x + 4)(x – 3) = 0
Check (x + 4)(x – 3 ) = 0
(–4 + 4)(–4 –3) 0
(0)(–7)
0
0
0
Check (x + 4)(x – 3 ) = 0
(3 + 4)(3 – 3)
(7)(0)
0
0
0
0
Substitute each
solution for x in
the original
equation.
9-5
Solving Quadratic Equations by Factoring
You may need to factor before using the Zero
Product Property. You can check your answers by
substituting into the original equation or by
graphing. If the factored form of the equation has
two different factors, the graph of the related
function will cross the x-axis in two places. If the
factored form has two identical factors, the graph
will cross the x-axis in one place.
9-5
Solving Quadratic Equations by Factoring
To review factoring techniques, see Lessons 8-3
through 8-5.
9-5
Solving Quadratic Equations by Factoring
Equations by Factoring
Solve the quadratic equation by factoring.
x2 – 6x + 8 = 0
(x – 4)(x – 2) = 0
x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
Check
x2 – 6x + 8 =
(4)2 – 6(4) + 8
16 – 24 + 8
0
0
0
0
0
Factor the trinomial.
Use the Zero Product
Property.
Solve each equation.
Check
x2 – 6x + 8 = 0
(2)2 – 6(2) + 8
4 – 12 + 8
0
0
0
0
9-5
Solving Quadratic Equations by Factoring
Equations by Factoring
Solve the quadratic equation by factoring.
x2 + 4x = 21
The equation must be written in
2
x + 4x = 21
standard form. So subtract
–21 –21
21 from both sides.
2
x + 4x – 21 = 0
(x + 7)(x –3) = 0
Factor the trinomial.
x + 7 = 0 or x – 3 = 0 Use the Zero Product Property.
x = –7 or x = 3
Solve each equation.
9-5
Solving Quadratic Equations by Factoring
Additional Example 2B Continued
Check Graph the related quadratic function. Because
there are two solutions found by factoring, the graph
should cross the x-axis in two places.
●
●
The graph of y = x2 + 4x – 21
intersects the x-axis at x = –7
and x = 3, the same as the
solutions from factoring. 
9-5
Solving Quadratic Equations by Factoring
Equations by Factoring
Solve the quadratic equation by factoring.
x2 – 12x + 36 = 0
(x – 6)(x – 6) = 0
x – 6 = 0 or x – 6 = 0
x=6
or
x=6
Factor the trinomial.
Use the Zero Product Property.
Solve each equation.
Both factors result in the same solution, so there
is one solution, 6.
9-5
Solving Quadratic Equations by Factoring
Additional Example 2C Continued
Check Graph the related quadratic function.
The graph of y = x2 – 12x + 36
shows one zero at 6, the same
as the solution from factoring. 
●
9-5
Solving Quadratic Equations by Factoring
Equations by Factoring
Solve the quadratic equation by factoring.
–2x2 = 20x + 50
The equation must be written in
–2x2 = 20x + 50
+2x2 +2x2
standard form. So add 2x2 to
0 = 2x2 + 20x + 50
both sides.
2x2 + 20x + 50 = 0
Factor out the GCF 2.
2(x2 + 10x + 25) = 0
2(x + 5)(x + 5) = 0
2≠0
or
x+5=0
Factor the trinomial.
Use the Zero Product Property.
x = –5 Solve the equation.
9-5
Solving Quadratic Equations by Factoring
Additional Example 2D Continued
Solve the quadratic equation by factoring.
–2x2 = 20x + 50
Check
–2x2 = 20x + 50
–2(–5)2
–50
–50
20(–5) + 50
–100 + 50
–50 
Substitute –5 into the
original equation.
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 2a
Solve the quadratic equation by factoring.
x2 – 6x + 9 = 0
(x – 3)(x – 3) = 0
x – 3 = 0 or x – 3 = 0
x = 3 or x = 3
The only solution is 3.
Check
x2 – 6x + 9 = 0
(3)2 – 6(3) + 9
9 – 18 + 9
0
0
0
0
Factor the trinomial.
Use the Zero Product
Property.
Solve each equation.
Substitute 3 into the
original equation.
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 2b
Solve the quadratic equation by factoring.
x2 + 4x = 5
x2 + 4x = 5
–5 –5
x2 + 4x – 5 = 0
(x – 1)(x + 5) = 0
x – 1 = 0 or x + 5 = 0
x=1
or
x = –5
Write the equation in standard
form. Add –5 to both sides.
Factor the trinomial.
Use the Zero Product Property.
Solve each equation.
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 2b Continued
Check Graph the related quadratic function. Because
there are two solutions found by factoring, the graph
should cross the x-axis in two places.
●
●
The graph of y = x2 + 4x – 5
intersects the x-axis at x = 1
and x = –5, the same as the
solutions from factoring.
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 2c
Solve the quadratic equation by factoring.
30x = –9x2 – 25
–9x2 – 30x – 25 = 0
–1(9x2 + 30x + 25) = 0
Write the equation in standard
form.
Factor out the GCF, –1.
–1(3x + 5)(3x + 5) = 0
Factor the trinomial.
–1 ≠ 0
Use the Zero Product Property.
– 1 cannot equal 0.
or
3x + 5 = 0
Solve the remaining equation.
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 2c Continued
Check Graph the related quadratic function.
Because there is one solution found by factoring,
the graph should cross the x-axis in one place.
●
The graph of y = –9x2 – 30x – 25
intersects the x-axis at x =
,
the same as the solution from
factoring.

9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 2d
Solve the quadratic equation by factoring. Check
3x2 – 4x + 1 = 0
(3x – 1)(x – 1) = 0
3x – 1 = 0 or x – 1 = 0
or x = 1
Factor the trinomial.
Use the Zero Product Property.
Solve each equation.
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 2d Continued
Check
3x2 – 4x + 1 = 0
3
–4
+1
0
0
0
Check
3x2 – 4x + 1 =
3(1)2 – 4(1) + 1
3–4+1
0
0
0
0
0
9-5
Solving Quadratic Equations by Factoring
Additional Example 3: Application
The height in feet of a diver above the water
can be modeled by h(t) = –16t2 + 8t + 8,
where t is time in seconds after the diver
jumps off a platform. Find the time it takes
for the diver to reach the water.
h = –16t2 + 8t + 8
0 = –8(2t2 – t – 1)
The diver reaches the water
when h = 0.
Factor out the GCF, –8.
0 = –8(2t + 1)(t – 1)
Factor the trinomial.
0=
–16t2
+ 8t + 8
9-5
Solving Quadratic Equations by Factoring
Additional Example 3 Continued
Use the Zero Product
–8 ≠ 0, 2t + 1 = 0 or t – 1= 0
Property.
2t = –1 or t = 1

It takes the diver 1 second
to reach the water.
Solve each equation.
Since time cannot be
negative,
does not
make sense in this
situation.
Check 0 = –16t2 + 8t + 8
0
–16(1)2 + 8(1) + 8 Substitute 1 into the
original equation.
0
–16 + 8 + 8
0
0
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 3
What if…? The equation for the height above
the water for another diver can be modeled
by h = –16t2 + 8t + 24. Find the time it takes
this diver to reach the water.
h = –16t2 + 8t + 24
0=
–16t2
+ 8t + 24
The diver reaches the water
when h = 0.
0 = –8(2t2 – t – 3)
Factor out the GCF, –8.
0 = –8(2t – 3)(t + 1)
Factor the trinomial.
9-5
Solving Quadratic Equations by Factoring
Check It Out! Example 3 Continued
–8 ≠ 0, 2t – 3 = 0 or t + 1= 0
2t = 3 or t = –1
t = 1.5
It takes the diver 1.5
seconds to reach the water.
Use the Zero Product
Property.
Solve each equation.
Since time cannot be
negative, –1 does not
make sense in this
situation.
Check 0 = –16t2 + 8t + 24
0
–16(1.5)2 + 8(1.5) + 24
Substitute 1.5 into the
0
–36 + 12 + 24
original equation.
0
0
9-5
Solving Quadratic Equations by Factoring
Lesson Quiz: Part I
Use the Zero Product Property to solve each
1. (x – 10)(x + 5) = 0 10, –5
2. (x + 5)(x) = 0
–5, 0
Solve each quadratic equation by factoring.
3. x2 + 16x + 48 = 0 –4, –12
4. x2 – 11x = –24 3, 8
9-5
Solving Quadratic Equations by Factoring
Lesson Quiz: Part II
5. 2x2 + 12x – 14 = 0 1, –7
6. x2 + 18x + 81 = 0
7. –4x2 = 16x + 16
–9
–2
8. The height of a rocket launched upward
from a 160 foot cliff is modeled by the
function h(t) = –16t2 + 48t + 160, where h
is height in feet and t is time in seconds.
Find the time it takes the rocket to reach the
ground at the bottom of the cliff.
5s
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