16 Days
Two Days

Review - Use FOIL and the Distributive
Property to multiply polynomials.
( x  3)( x  5) 
(2 x  1)( x  5) 



Writing a polynomial as a product of its
factors.
Essentially undoing the multiplication.
Purposes of factoring:
◦ Simplifying
◦ Rewriting
◦ Solving
x  10 x  24 
2
x 16 
2

GCF

Grouping

Difference of Squares

Difference of Squares Formula
a  b  (a  b)( a  b)
2
2

Perfect Square Trinomials
x  6x  9 
2
4 x  4 xy  y 
2
2

Perfect Square Trinomial Formulas
a  2ab  b  (a  b)
2
a  2ab  b  (a  b)
2
2
2
2
2

Trinomials (a=1)

Trinomials (a≠1)
2 x  5x  3 
2
6x  x  2 
2

Sum and Difference of Cubes Formulas
x  y  ( x  y)( x  2 xy  y )
3
3
2
2
x  y  ( x  y)( x  2 xy  y )
3
3
2
2

Sum and Difference of Cubes

pg 263 (# 2-30 even)
x  7x  6
2
x  16
2
x  3 x  18
2
x  11x  18
2
2x  9x  4
2
x  9x
3
2x  6x  4
2
10 x  13 x  3
2
6 x  5x  6
2
16 x  25
2
Two Days

Question: How can we multiply two or more numbers
together and get a product that equals zero?

For any real numbers a and b, if ab=0, then either
a=0, b=0, or both.
x  32x 1  0
x  3  0 or 2x  1  0
x  3 or 2x  1
1
x  3 or x 
2
The solutions are -3 and 1/2.
2
9x  4  0
Factor using “difference of two squares.”
3x  23x  2   0
3x  2  0 or 3x  2  0
3x  2 or 3x  2
2
2
x   or x 
3
3
2
x  27  6x
2
x  6x  27  0
x  9x  3  0
x  9  0 or x  3  0
x  9 or x  3
2
2x 8x  0
2x x  4   0
2x  0 or x  4  0
x  0 or x  4
Did you take out GCF?
1. 3y  52y  7  0
2
2. x  x  12
2
3. d  5d  0
4. 4c  25
2
5. 18u  1  3u
2





1. y  5 3 or y  7 2
2. x  4 or x  3
3.
d  0 or d  5
4.
5. c  5 2 or c  5 2
u  1 6 or u  1 3
If all are correct,
you’re finished!
3y  52y  7  0
3y  5  0 or 2y  7  0
3y  5 or 2y  7
y  5 3 or y   7 2
Back to questions
2
x  x  12
2
x  x  12  0
x  4 x  3  0
x  4  0 or x  3  0
x  4 or x  3
Back to questions
d  5d  0
2
d d  5  0
d  0 or d  5  0
d  0 or d  5
Back to questions
4c2  25
4c  25  0
2
2c  52c  5  0
2c  5  0 or 2c  5  0
2c  5 or 2c  5
c   5 2 or c  5 2
Back to questions
2
18u  3u  1
2
18u  3u  1  0
6u  13u  1  0
6u  1  0 or 3u  1  0
6u  1or 3u  1
u  1 6 or u  1 3
Back to questions
One Day
Two Days
Solve the following quadratic:
x2  4  0




The Imaginary Unit (i) has the following
properties.
The imaginary number i is defined as the
number whose square is -1. That is:
Imaginary Numbers are of the form a + bi
where b ≠ 0.
Complex Numbers are of the form a + bi
where a and b are Real Numbers.

We can add and subtract imaginary numbers
similar to how we add and subtract terms
with variables. Think “like terms.”

Similarly, we can multiply imaginary numbers
following the same exponent rules we use for
variables.


The absolute value of a complex number is
the distance the number lies from the origin
in the complex plane.
Think Pythagorean Thm..
a  bi  a 2  b2

Larger powers of i can be simplified by
dividing the power by 4 and using the
remainder to determine the appropriate
value.

Solve:
5 x 2  200  0

Pg 278 (# 1-45 odd)


If z = a + bi is an imaginary number, the its
conjugate is z = a – bi.
Complex Conjugates can be used to eliminate
imaginary numbers from the denominators of
fractions. This is very similar to how we
rationalize denominators.

Eliminate the Imaginary numbers from the
denominator in the following example.

Practice 5-6 WS (even)
Two Days
 Examples
 x2 + 6x +
9
 x2 - 10x + 25
 x2 + 12x + 36
 In
the following perfect square
trinomial, the constant term is
missing.
X2 + 14x + ____
 Find the constant term by
squaring half the coefficient of
the linear term.
 (14/2)2
X2 + 14x + 49
ax 2  bx  c  0
 b  b 2  4ac
Quardatic Formula : x 
2a

Solve the following using the quadratic
formula:
x2  6x  5  0
2 x 2  3x  9  0
What do we notice about these two problems? How else could we
Have solved these quadratics?

Solve the following quadratics:
3x 2  10 x  5  0
3 x 2  4 x  10  0

pg 293 (# 1-29 odd)
1 Day
1. When you solve using completing the square
on the general formula ax 2  bx  c  0
you get:
b  b2  4ac
x
2a
2. This is the quadratic formula!
3. Just identify a, b, and c then substitute into
the formula.
The quadratic formula allows you to solve
ANY quadratic equation, even if you
cannot factor it.
An important piece of the quadratic formula
is what’s under the radical:
b2 – 4ac
This piece is called the discriminant.
The discriminant tells you the number and types of answers
(roots) you will get. The discriminant can be +, –, or 0
which actually tells you a lot! Since the discriminant is
under a radical, think about what it means if you have a
positive or negative number or 0 under the radical.
Value of the Discriminant
Nature of the Solutions
Negative
2 imaginary solutions
Zero
1 Real Solution
Positive – perfect square
2 Reals- Rational
Positive – non-perfect
square
2 Reals- Irrational
x2  4x  5  0
2 x 2  x  28  0
6x2  2x  5  0
x 2  12 x  36  0

Pg 293 (#31-39 odd)
Practice 5-8 WS (#2-26 even)

Quiz 5.8 on 11/18!!

1 Day

The standard form of a quadratic is
y  ax  bx  c where :
2
2
ax is the quadratic term,
bx is the linear ter m, and
c is the constant t erm.



The graph of a quadratic function is called a
parabola.
The axis of symmetry is the vertical line that
divides the parabola into two identical halves
and is written x=a.
The vertex (a,b) of the parabola is the point
at which the parabola intersects the axis of
symmetry and is also a maximum or
minimum point of the function.

Given 3 points on the function we can
determine the equation of the quadratic.
Find y  ax  bx  c where :
(1,0), (2,3), and (3,10) are points on the function.
2

pg. 241 (#1-12 all, 21)
4 Days
3 Days

Standard form of a Quadratic:
y  ax  bx  c ; a  0
2

Vertex form of a Quadratic:
y  a( x  h) 2  k ; Vertex : (h, k )
y  3x 2  6 x  4
y  3( x 2  2 x 
)4
y  3( x 2  2 x  (1) 2 )  4  3(1) 2
y  3( x  1) 2  7 Vertex : (-1,-7), a  3