Chapter 6
Factoring
Chapter Sections
6.1 – Factoring a Monomial from a Polynomial
6.2 – Factoring by Grouping
6.3 – Factoring Trinomials of the Form
ax2 + bx + c, a = 1
6.4 – Factoring Trinomials of the Form
ax2 + bx + c, a ≠ 1
6.5 – Special Factoring Formulas and a General Review
of Factoring
6.6 – Solving Quadratic Equations Using Factoring
6.7 – Applications of Quadratic Equations
Angel, Elementary and Intermediate Algebra, 3ed 2
§ 6.1
Factoring a
Monomial from a
Polynomial
Factors
To factor an expression means to write the
expression as a product of its factors.
If a · b = c, then a and b are factors of c.
a·b
Recall that the greatest common factor (GCF) of
two or more numbers is the greatest number that will
divide (without remainder) into all the numbers.
Example: The GCF of 27 and 45 is 9.
Angel, Elementary and Intermediate Algebra, 3ed 4
Factors
A prime number is an integer greater than 1 that
has exactly two factors, 1 and itself.
A composite number is a positive integer that
is not prime.
Prime factorization is used to write a number
as a product of its primes.
24 = 2 · 2 · 2 · 3
Angel, Elementary and Intermediate Algebra, 3ed 5
Determining the GCF
1. Write each number as a product of prime
factors.
2. Determine the prime factors common to all
the numbers.
3. Multiply the common factors found in step 2.
The product of these factors is the GCF.
Example: Determine the GCF of 24 and 30.
24 = 2 · 2 · 2 · 3 30 = 2 · 3 · 5
A factor of 2 and a factor of 3 are common
to both, therefore 2 · 3 = 6 is the GCF.
Angel, Elementary and Intermediate Algebra, 3ed 6
Determining the GCF
To determine the GCF of two or more terms, take
each factor the largest number of times it appears in
all of the terms.
Example:
a.) The GCF of 6p, 4p2 and 8p3 is 2p.
2·3·p
2·2·p·p 2·2 ·2·p·p·p
b.) The GCF of 4x2y2, 3xy4, and 2xy2 is xy2.
Angel, Elementary and Intermediate Algebra, 3ed 7
Factoring Monomials from Polynomials
1. Determine the GCF of all the terms in the
polynomial.
2. Write each term as the product of the GCF
and its other factors.
3. Use the distributive property to factor out
the GCF.
Example: 24x + 16x3 (GCF is 8x)
= 8·3·x + 8·2·x·x2 = 8x (3 + 2x2)
(To check, multiply the factors using the distributive property. )
Angel, Elementary and Intermediate Algebra, 3ed 8
§ 6.2
Factoring by
Grouping
Factoring by Grouping
The process of factoring a polynomial containing
four or more terms by removing common factors
from groups of terms is called factoring by
grouping.
Example: Factor x2 + 7x + 3x + 21.
x(x + 7) + 3(x + 7) =
(x + 7) (x + 3)
(Use the FOIL method to check your answer.)
Angel, Elementary and Intermediate Algebra, 3ed 10
Factoring by Grouping
1. Determine whether there are any factors
common to all four terms. If so, factor the
GCF from each of the four terms.
2. If necessary, arrange the four terms so that the
first two terms have a common factor and the
last tow terms have a common factor.
3. Use the distributive property to factor each
group of two terms.
4. Factor the GCF from the results of step 3.
Angel, Elementary and Intermediate Algebra, 3ed 11
Factoring by Grouping
Example:
a.) Factor by grouping: 4x2 – 6x + 6x – 9
2x(2x – 3) + 3(2x – 3) =
(2x + 3) (2x – 3)
b.) Factor by grouping: 5x2 + 20x – x – 4
5x(x + 4) + (– 1) (x + 4) =
(5x – 1) (x + 4)
Angel, Elementary and Intermediate Algebra, 3ed 12
§ 6.3
Factoring Trinomials
of the Form
ax2 + bx + c, a = 1
Factoring Trinomials
Recall that factoring is the reverse process of
multiplication. Using the FOIL method, we can
show that
(x – 3)(x – 8) = x2 – 11x + 24.
Therefore x2 – 11x + 24 = (x – 3)(x – 8).
Note that this trinomial results in the product
of two binomials whose first term is x and
second term is a number (including its sign).
Angel, Elementary and Intermediate Algebra, 3ed 14
Factoring Trinomials
Factoring any polynomial of the form x2 + bx + c
will result in a pair of binomials:
x2 + bx + c = (x +?)(x +?)
Numbers go here.
L
F
F O I L
(7x + 3)(2x + 4) = 14x2 + 28x + 6x + 12
I = 14x2 + 34x + 12
O
Angel, Elementary and Intermediate Algebra, 3ed 15
Factoring Trinomials
1. Find two numbers whose product equals
the constant, c, and whose sum equals the
coefficient of the x-term, b.
2. Use the two numbers found in step 1,
including their signs, to write the trinomial
in factored form. The trinomial in factored
form will be
(x + one number) (x + second number)
Angel, Elementary and Intermediate Algebra, 3ed 16
Examples
a.) Factor x2 + 8x + 15.
x2 + 8x + 15 = (x + ?) (x + ?)
Replace the ?s with two numbers that are the
product of 15 and the sum of 8.
x2 + 8x + 15 = (x + 3) (x + 5)
b.) Factor x2 – 10x – 25.
This is a prime polynomial because it
cannot be factored.
Angel, Elementary and Intermediate Algebra, 3ed 17
Examples Continued
c.) Factor 2a2 – 12a – 32.
First, factor out a 2 from the polynomial.
2a2 – 12a – 32 = 2(a2 – 6a – 16)
Then factor the trinomial.
2(a2 – 6a – 16) = 2(a – 8)(a + 2)
d.) Factor 2b5 + 16b4 + 30b3.
2b5 + 16b4 + 30b3 = 2b3(b2 + 8b + 15)
= 2b3(b + 3)(b + 5)
Angel, Elementary and Intermediate Algebra, 3ed 18
§ 6.4
Factoring Trinomials
of the Form
ax2 + bx + c, a ≠ 1
Trial and Error Method
1. Determine whether there is any factor common
to all three terms. If so, factor it out.
2. Write all pairs of factors of the coefficient of the
squared term, a.
3. Write all pairs of factors of the constant term, c.
4. Try various combinations of these factors until
the correct middle term, bx, is found.
Angel, Elementary and Intermediate Algebra, 3ed 20
Trial and Error Method
Example: Factor 2x2 + 9x + 4.
There is no factor common to all three terms.
Since the first term is 2x2, one factor must
contain 2x and the other an x.
(2x + ?)(x + ?)
The product of the last term in the factors
must be 4. Only the positive factors of 12
will be considered.
Angel, Elementary and Intermediate Algebra, 3ed 21
Trial and Error Method
Factor 2x2 + 9x + 4.
Sum of the Products
Possible Factors
Factors of 4 of the Inner and
of Trinomial
Outer Terms
1(4) (2x + 1)(x + 4) 9x
2(2) (2x + 2)(x + 2) 6x
4(1) (2x + 4)(x + 1) 6x
Since the product of (2x + 1)(x + 4) yields the
correct term, 9x, they are the correct factors.
2x2 + 9x + 4 = (2x + 1)(x + 4)
Angel, Elementary and Intermediate Algebra, 3ed 22
Factor by Grouping Method
1. Determine whether there is any factor common to
all three terms. If so, factor it out.
2. Find two numbers whose product is equal to the
product of a times c, and whose sum is equal to b.
3. Rewrite the middle term, bx, as the sum or
difference of two terms using the numbers found
in step 2.
4. Factor by grouping.
Angel, Elementary and Intermediate Algebra, 3ed 23
Factor by Grouping Method
Example: Factor 2x2 + 9x + 4.
There is no factor common to all three
terms.
a=2 b=9 c=4
Find two numbers whose product is a · c and
whose sum is b.
Factors of Sum of Factors
8 9
(1)(8) 6
(2)(4) Continued.
Angel, Elementary and Intermediate Algebra, 3ed 24
Factor by Grouping Method
Example continued:
Factor 2x2 + 9x + 4.
Use these factors to rewrite 9x.
2x2 + 9x + 4
2x2 + 1x + 8x + 4
Factor by grouping.
2x2 + 1x + 8x + 4 =
x (2x + 1) + 4 (2x + 1) = FOIL to check.
(2x + 1) (x + 4)
Angel, Elementary and Intermediate Algebra, 3ed 25
Factor by Grouping Method
Example: Factor 18x3 – 21x2 – 9x.
Factor out 3x.
18x3 – 21x2 – 9x = 3x (6x2 – 7x – 3)
Rewrite the middle term.
3x (6x2 – 7x – 3) = 3x (6x2 – 9x + 2x – 3)
Factor by grouping.
3x (6x2 – 9x + 2x – 3) = 3x[3x(2x – 3) +1 (2x – 3)]
= 3x (3x + 1) (2x – 3)
FOIL to check.
Angel, Elementary and Intermediate Algebra, 3ed 26
§ 6.5
Special Factoring
Formulas and a General
Review of Factoring
Difference of Two Squares
a2 – b2 = (a + b) (a – b)
Example:
a.) Factor x2 – 49.
x2 – 49 = x2 – 72 = (x + 7)(x – 7)
b.) Factor 4x4 – 25y8.
4x4 – 25y8 = (2x2)2 – (5y4)2 =
(2x2 + 5y4)(2x2 – 5y4)
Angel, Elementary and Intermediate Algebra, 3ed 28
Sum of Two Cubes
a3 + b3 = (a + b) (a2 – ab + b2)
Example:
a.) Factor k3 + 8.
k3 + 8 = k3 + 23 = (k + 2)(k2 – 2k + 4)
b.) Factor 27c3 + 125d3.
(a = 3c ; b = 5d)
27c3 + 125d3 = (3c + 5d)(9c2 – 15cd + 25d2)
Angel, Elementary and Intermediate Algebra, 3ed 29
Difference of Two Cubes
a3 – b3 = (a – b) (a2 + ab + b2)
Example:
a.) Factor p3 – 1.
p3 – 1 = (p – 1)(p2 + p + 1)
b.) Factor 64x3 – 125y3.
(a = 4x ; b = 5y)
64x3 – 125y3 = (4x – 5y)(16x2 + 20xy + 25y2)
Angel, Elementary and Intermediate Algebra, 3ed 30
Helpful Hint for Factoring
When factoring the sum or difference of two cubes, the
sign between the terms in the binomial factor will be the
same as the sign between the terms.
The sign of the ab term will be the opposite of the sign
between the terms of the binomial factor.
The last term in the trinomial will always be positive.
a3 + b3 = (a + b) (a2 – ab + b2)
same sign
opposite sign always positive
a3 – b3 = (a – b) (a2 + ab + b2)
same sign
opposite sign always positive
Angel, Elementary and Intermediate Algebra, 3ed 31
§ 6.6
Solving Quadratic
Equations Using
Factoring
Quadratic Equation
A quadratic equation is an equation
that contains a second-degree term and
no term of a higher degree.
Quadratic equations have the form
ax2 + bx + c = 0
where a, b, and c are real numbers, a 0
Examples:
a.) z2 + 3z + 7 = 0
b.) 4k2 – 5 = 0
Angel, Elementary and Intermediate Algebra, 3ed 33
Zero-Factor Property
To solve a quadratic equation by
factoring, the zero-factor property is
used.
If ab = 0, then a = 0 or b = 0
Example:
Solve the equation 3x(x + 4) = 0
3x = 0 or x+ 4 = 0
x=0 x = -4
Angel, Elementary and Intermediate Algebra, 3ed 34
Solving with Factoring
1. Write the equation in standard form with
the standard term having a positive
coefficient. This will result in one side of
the equation being 0.
2. Factor the side of the equation that is not 0.
3. Set each factor containing a variable equal to
0 and solve each equation.
4. Check each solution found in step 3 in the
original equation.
Angel, Elementary and Intermediate Algebra, 3ed 35
Solving with Factoring
Example: Solve the following equations.
a.) –9x + 20 = -x2
x2 – 9x + 20 = 0
(x – 4) (x – 5) = 0
x – 4 = 0 or x – 5 = 0
x=4 or x = 5
Check:
– 9(4) + 20 = – (4)2 – 36 + 20 = – 16
– 9(5) + 20 = – (5)2 – 45 + 20 = – 25
Angel, Elementary and Intermediate Algebra, 3ed 36
Solving with Factoring
b.) 2x2 = 50
2x2 – 50 = 0
2(x2 – 25) = 0
2(x – 5) (x + 5) = 0
x – 5 = 0 or x + 5 = 0
x = 5 or x = – 5
Check:
2(5)2 = 50 2(25) = 50
2(-5)2 = 50 2(25) = 50
Angel, Elementary and Intermediate Algebra, 3ed 37
§ 6.7
Applications of
Quadratic Equations
Solving Applications
Example:
The area of a rectangle is 84 square
inches. Determine the length and width
if the length is 2 inches less than twice
the width.
A = lw = 84 w=? l = 2w – 2
A = 84
l=?
lw = 84
(2w – 2)w = 84
Solve.
Continued.
Angel, Elementary and Intermediate Algebra, 3ed 39
Solving Applications
Example continued:
(2w – 2)w = 84
2w2 – 2w = 84
2w2 – 2w – 84 = 0
2(w2 – w – 42) = 0
2(w – 7) (w + 6) = 0
w – 7 = 0 or w + 6 = 0
w = 7 or w = -6
The width of the rectangle is 7 inches and
the length is 2(7) – 2 = 14 inches.
Angel, Elementary and Intermediate Algebra, 3ed 40
Pythagorean Theorem
The square of the hypotenuse of a right
triangle is equal to the sum of the squares of
the two legs.
(leg)2 + (leg)2 = (hypotenuse)2
If a and b represent the legs, and c represents
the hypotenuse, then
a2 + b2 = c2
c b
a
Angel, Elementary and Intermediate Algebra, 3ed 41
Pythagorean Theorem
Example:
One leg of a right triangle is two inches
more than twice the other leg. The
hypotenuse is 13 inches. Find the length
of the three sides of the triangle.
Let a = length of the first leg
13 b Let b = length of the second leg
a b = 2 + 2a
a2 + b2 = c2
a2 + (2 + 2a)2 = 132 Solve.
Continued.
Angel, Elementary and Intermediate Algebra, 3ed 42
Pythagorean Theorem
Example continued:
a2 + b2 = c2
a2 + (2 + 2a)2 = 132
a2 + 4 + 8a + 4a2 = 169
5a2 + 8a + 4 = 169
5a2 + 8a – 165 = 0
5a2 – 25a + 33a – 165 = 0
5a(a – 5) + 33(a – 5) = 0
(5a + 33)(a – 5) = 0
5a + 33 = 0 or a – 5 = 0
a = -33/5 or a = 5
The lengths of the triangle are 5 in., 12 in., and 13 in.
Angel, Elementary and Intermediate Algebra, 3ed 43