Chapter 5
Algebra IIA
K. Audette
2010- 2011
GCF
2 terms
a
2
a
3
b
b
2
3 - terms
4 -terms
ax 2 bx c
grouping
3
a 3 b3
Repeat if necessary
Reading Math
• Functions have zeros or x-intercepts.
• Equations have solutions or roots.
Check It Out! Example
Find the zeros of the function by factoring.
g(x) = x2 – 8x
x2 – 8x = 0
x(x – 8) = 0
x = 0 or x – 8 = 0
x = 0 or x = 8
Set the function to equal to 0.
Factor: The GCF is x.
Apply the Zero Product Property.
Solve each equation.
Example 2A: Finding Zeros by Factoring
Find the zeros of the function by factoring.
f(x) = x2 – 4x – 12
x2 – 4x – 12 = 0
(x + 2)(x – 6) = 0
x + 2 = 0 or x – 6 = 0
x= –2 or x = 6
Set the function equal to 0.
Factor: Find factors of –12 that add to –4.
Apply the Zero Product Property.
Solve each equation.
Example 2B: Finding Zeros by Factoring
Find the zeros of the function by factoring.
g(x) = 3x2 + 18x
3x2 + 18x = 0
3x(x+6) = 0
3x = 0 or x + 6 = 0
x = 0 or x = –6
Set the function to equal to 0.
Factor: The GCF is 3x.
Apply the Zero Product Property.
Solve each equation.
Check It Out! Example 2a
Find the zeros of the function by factoring.
f(x)= x2 – 5x – 6
x2 – 5x – 6 = 0
(x + 1)(x – 6) = 0
x + 1 = 0 or x – 6 = 0
x = –1 or x = 6
Set the function equal to 0.
Factor: Find factors of –6 that add to –5.
Apply the Zero Product Property.
Solve each equation.
Quadratic expressions can have one, two or three
terms, such as –16t2, –16t2 + 25t, or –16t2 + 25t + 2.
Quadratic expressions with two terms are binomials.
Quadratic expressions with three terms are trinomials.
Some quadratic expressions with perfect squares have
special factoring rules.
Example 4A: Find Roots by Using Special Factors
Find the roots of the equation by factoring.
4x2 = 25
4x2 – 25 = 0
(2x)2 – (5)2 = 0
Rewrite in standard form.
Write the left side as a2 – b2.
(2x + 5)(2x – 5) = 0
Factor the difference of squares.
2x + 5 = 0 or 2x – 5 = 0
Apply the Zero Product Property.
x=–
or x =
Solve each equation.
Example 4 Continued
Check Graph the related function f(x) = 4x2 – 25 on
a graphing calculator. The function appears to
have zeros at
and .
10
–3
55
3
30
Example 4B: Find Roots by Using Special Factors
Find the roots of the equation by factoring.
18x2 = 48x – 32
18x2 – 48x + 32 = 0
2(9x2 – 24x + 16) = 0
9x2 – 24x + 16 = 0
(3x)2 – 2(3x)(4) + (4)2 = 0
(3x – 4)2 = 0
3x – 4 = 0 or 3x – 4 = 0
x=
or x =
Rewrite in standard form.
Factor. The GCF is 2.
Divide both sides by 2.
Write the left side as a2 – 2ab +b2.
Factor the perfect-square trinomial.
Apply the Zero Product Property.
Solve each equation.
Check It Out! Example 4a
Find the roots of the equation by factoring.
x2 – 4x = –4
x2 – 4x + 4 = 0
(x – 2)(x – 2) = 0
x – 2 = 0 or x – 2 = 0
x = 2 or x = 2
Rewrite in standard form.
Factor the perfect-square trinomial.
Apply the Zero Product Property.
Solve each equation.
Check It Out! Example 4b
Find the roots of the equation by factoring.
25x2 = 9
25x2 – 9 = 0
(5x)2 – (3)2 = 0
Rewrite in standard form.
Write the left side as a2 – b2.
(5x + 3)(5x – 3) = 0
Factor the difference of squares.
5x + 3 = 0 or 5x – 3 = 0
Apply the Zero Product Property.
x=
or x =
Solve each equation.
Check It Out! Example 4b Continued
Check Graph the related function f(x) = 25x2 – 9
on a graphing calculator. The function appears to
have zeros at
and .
10
1
–1
10
Example 2: Factoring by Grouping
Factor: x3 – x2 – 25x + 25.
(x3 – x2) + (–25x + 25)
Group terms.
x2(x – 1) – 25(x – 1)
Factor common monomials from each
group.
(x – 1)(x2 – 25)
Factor out the common binomial (x –
1).
(x – 1)(x – 5)(x + 5)
Factor the difference of squares.
Check It Out! Example 2a
Factor: x3 – 2x2 – 9x + 18.
(x3 – 2x2) + (–9x + 18)
Group terms.
x2(x – 2) – 9(x – 2)
Factor common monomials from each
group.
(x – 2)(x2 – 9)
Factor out the common binomial (x –
2).
(x – 2)(x – 3)(x + 3)
Factor the difference of squares.
Check It Out! Example 2b
Factor: 2x3 + x2 + 8x + 4.
(2x3 + x2) + (8x + 4)
Group terms.
x2(2x + 1) + 4(2x + 1)
Factor common monomials from each
group.
(2x + 1)(x2 + 4)
Factor out the common binomial (2x +
1).
(2x + 1)(x2 + 4)
Just as there is a special rule for factoring the difference of two squares, there are special
rules for factoring the sum or difference of two cubes.
Example 3A: Factoring the Sum or Difference of Two Cubes
Factor the expression.
4x4 + 108x
4x(x3 + 27)
Factor out the GCF, 4x.
4x(x3 + 33)
Rewrite as the sum of cubes.
4x(x + 3)(x2 – x 3 + 32)
Use the rule a3 + b3 = (a + b)
+ b2).
4x(x + 3)(x2 – 3x + 9)
(a2 – ab
Example 3B: Factoring the Sum or Difference of Two Cubes
Factor the expression.
125d3 – 8
(5d)3 – 23
(5d – 2)[(5d)2 + 5d 2 + 22]
(5d – 2)(25d2 + 10d + 4)
Rewrite as the difference of cubes.
Use the rule a3 – b3 =
(a – b) (a2 + ab + b2).
Check It Out! Example 3a
Factor the expression.
8 + z6
(2)3 + (z2)3
(2 + z2)[(2)2 – 2 z + (z2)2]
(2 + z2)(4 – 2z + z4)
Rewrite as the difference of cubes.
Use the rule a3 + b3 =
(a + b) (a2 – ab + b2).
Check It Out! Example 3b
Factor the expression.
2x5 – 16x2
2x2(x3 – 8)
2x2(x3 – 23)
2x2(x – 2)(x2 + x 2 + 22)
2x2(x – 2)(x2 + 2x + 4)
Factor out the GCF, 2x2.
Rewrite as the difference of cubes.
Use the rule a3 – b3 = (a – b)
+ b2).
(a2 + ab