MAT0024-3
Factoring Polynomials
Purpose
The purpose of this handout is to help the student recognize different types of
polynomials, and then to apply the appropriate strategy to factor them.
Overview
3 Terms (trinomial)
• special case (perfect
square trinomial)
• ax2 + bx +c when a = 1
• ax2 + bx +c when a ≠ 1
1st
Factor
GCF
(if possible)
2 Terms (binomial)
• Special case (difference
of squares)
• Special case (difference
of cubes or sum of
cubes)
4 Terms
• Try factoring by
grouping
When factoring is completed:
a. Check to see if one of the polynomial factors can be factored again.
b. Double check for other GCFs (greatest common factors)
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GCF Greatest Common Factor
•
•
A factor that every term in the expression has in common
Reverse distribution
Example:
12 x 4 − 6 x 3 + 3 x 2
To find the GCF for
12 x 4 − 6 x 3 + 3 x 2
means ( 2 ⋅ 2 ⋅ 3 ⋅ x ⋅ x ⋅ x ⋅ x ) − ( 2 ⋅ 3 ⋅ x ⋅ x ⋅ x ) + (3 ⋅ x ⋅ x )
all three terms have 3 and x ⋅ x as common factors
( 2 ⋅ 2 ⋅ 3 ⋅ x ⋅ x ⋅ x ⋅ x ) − ( 2 ⋅ 3 ⋅ x ⋅ x ⋅ x ) + (3 ⋅ x ⋅ x )
2
So the GCF is 3x
Notes:
a. 3 is the greatest common numerical factor because it is the largest number that
divides into every term evenly.
b. x2 is the greatest common variable factor because it is the variable to the
largest power that divides evenly into all three terms.
To factor 12 x 4 − 6 x 3 + 3 x 2
think reverse distribution
3x2(4x2 – 2x + 1)
Factoring Trinomials (ax2 + bx + c)
1. Special Case (Perfect Square Trinomial)
Strategy: Test the trinomial to see if it satisfies the conditions of a Perfect Square
Trinomial.
a. A Perfect Square Trinomial:
A2 + 2AB + B2 factors into (A + B)(A + B) which can be written (A + B)2
or A2 – 2AB + B2 factors into (A – B)(A – B) which can be written (A – B)2
Test: 1) 1st and last terms must be perfect squares.
2) │middle term│ must equal 2 times the roots of the 1st and last
terms.
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Example:
Factor
9 x − 24 x + 16
2
Test: 1) 1st term is 9x2 → (3x)2 and last term is 16 → (4)2 , both
perfect squares with roots 3x and 4 as shown.
2) Does the│middle term│ 24x equal 2(3x)(4) ?
YES!
The trinomial passes both tests. It is a Perfect Square Trinomial of the type
A2 – 2AB + B2 = (A – B)2 where 3x → A and 4 → B
Solution:
(3 x − 4 )2
b. For a = 1 (lead coefficient is 1)
Strategy: Find the product a ⋅ c then determine which factor pairs of ac add up
to b.
Example:
Factor
x 2 − x − 12 Note: a = 1, b = -1, and c = -12
a ⋅ c ⇒ (1)( −12) = −12 = ac
List Factor Pairs of ac
1
-12
2
-6
Note: 1) You don’t have to list all factor pairs,
3
-4 adds to b ⇒ −1
just find the two that add to b.
4
-3
2) If there are no factor pairs that add to b
-1
12
then the polynomial is “prime”.
-2
6
-3
4
-4
3
Solution must be of the form (x )(x ) where the factor pair and corresponding
positive or negative sign supply the missing terms.
Solution:
(x + 3)(x − 4)
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1
a
c. For
≠
Strategy: Find the product a ⋅ c then determine the factor pairs of ac that add up
to b.
Example:
Factor 4 x 2 − 25 x − 21
Note: a = 4, b = -25, c = -21
a ⋅ c ⇒ ( 4)( −21) = −84 = ac
List Factor Pairs of ac
1
-84
2
-42
3
-28 adds to b ⇒ −25
4
-21
Note: 1) You don’t have to list all factor pairs,
just find the two that add to b.
2) If there are no factor pairs that add to b
Add to b then the polynomial is “prime”.
.
.
Solution must be of the form ( x
)( x
)
Clues to help you find the missing coefficients and terms:
1) List the factor pairs of a to find possible coefficients of x.
2) List the factor pairs of c to find possible missing final terms.
+3
iii) ( x
)( x ) Product of “inners” and “outers” must be from
the list of factor pairs of ac above that add to b.
-28
Solution:
(4 x + 3)(x − 7)
Factor by Grouping (4 Terms)
Strategy: Group the four terms in pairs and find the GCF of each pair. Both
pairs should then have a common binomial factor.
Example 1: Factor 3ax − 3ay − 2bx + 2by
3ax − 3ay − 2bx + 2by
GCF of 1st group is 3a
GCF of 2nd group is
− 2b
(Note: If the two groups are separated by a subtraction always factor out a
negative GCF from the second group.)
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MAT0024-3
Both have a common binomial factor (x – y) which can be factored out leaving
the other factor (3a – 2b).
Solution:
(x − y )(3a − 2b )
Example 2: Factor 4 x 2 − 25 x − 21
1
a
Strategy: Notice! This polynomial is the same as in the Section “Factoring
Trinomials” and it doesn’t have 4 terms. But, if we replace the middle term -25x
with the factor pair of ac from the list that adds to -25 we can write this trinomial
with 4 terms and solve by grouping. This is an alternative way of factoring a
trinomial where ≠ .
4 x 2 − 28 x + 3 x − 21
GCF of 1st group is 4 x
Factoring each group yields
GCF of 2nd group is 3
4 x( x − 7 ) + 3( x − 7)
Both have a common binomial factor (x – 7) which can be factored out leaving
the other factor (4x + 3)
Solution:
(x − 7)(4 x + 3)
Special Cases (2 Terms)
1. Difference of Squares
Strategy: Check the binomial to see if it is a difference of squares.
A Difference of Squares A2 – B2 factors into (A + B)(A – B)
TEST: a) 1st and last terms must be perfect squares.
b) Must be a subtraction.
NOTE: Besides a possible common factor, the Sum of Squares
A2 + B2 is PRIME (not factorable).
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Example:
9 x − 16
Factor
2
Test: a) 1st term is 9x2 → (3x)2 and last term is 16 → (4)2 ,
both perfect squares with roots 3x and 4 as shown.
b) Are we subtracting? YES!
The binomial passes both tests. It is a Difference of Squares of the form
A2 – B2 = (A + B)(A – B) where 3x → A and 4 → B
Solution:
2.
(3x + 4)(3x − 4)
Difference OR Sum of Cubes
Strategy: Check the binomial to see if it is a difference or a sum of cubes.
A Difference of Cubes
A Sum of Cubes
A3 – B3 factors into (A – B)(A2 + AB + B2)
A3 + B3 factors into (A + B)(A2 – AB + B2)
Note: The trinomial factor is often PRIME and cannot be factored further.
TEST: a) 1st and last terms must be perfect cubes.
b) Can be either a subtraction OR an addition.
Example 1: Factor 27 x 3 + 8
Test: a) 1st term is 27x3 → (3x)3 and last term is 8 → (2)3 ,
both perfect cubes with roots 3x and 2 as shown.
b) it is the sum of cubes.
The binomial is a Sum of Cubes of the form
A3 + B3 = (A + B)(A2 – AB + B2), where 3x → A and 2 → B
Solution:
(3x + 2)(9 x 2 − 6 x + 4)
Example 2: Factor 1 − 64 x 3
Test: a) 1st term is 1 → (1)3 and last term is 64x3 → (4x)3 ,
both perfect cubes with roots 1 and 4x as shown.
b) it is the difference of cubes.
The binomial is a Difference of Cubes of the form A3 – B3 =
(A – B)(A2 + AB + B2), where 1 → A and 4x → B
Solution:
(1 − 4 x )(1 + 4 x + 16 x 2 )
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WORKED EXAMPLES
1.
Factor 16x4 – 1
(4x2)2 – (1)2
(4x2 + 1)(4x2 – 1)
(4x2 + 1)((2x)2 – (1)2)
(4x2 + 1)(2x + 1)(2x – 1)
2.
3x2 + 27
Factor
3(x2 + 9)
3.
Difference of Squares
Another Difference of Squares!
GCF
(x2 + 9) can NOT be factored further it is
the Sum of Squares which is prime.
8a3 + 27b3
Factor
Sum of Cubes
(2a)3 + (3b)3
2
2
(2a + 3b)(4a – 6ab + 9b )
4.
Factor
( x
2x2 – xy – 6y2
y)( x
y)
+3
Factor pairs of ac = -12 that add to the middle
are 3 and -4
( 2 x + 3y)( 1 x – 2y)
-4
(2x + 3y)(x – 2y)
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Practice Exercises
1.
5x2 – 5
11.
x2(x – 2) – (x – 2)
2.
x2 + 3x – 10
12.
24x2y + 6x3y – 45x4y
3.
2x2 + 3x – 2
13.
a2b2 + 6ab2 + 9b2
4.
x3 +4x2 + 4x
14.
54a4 + 16a
5.
50 – 2x2
6.
4ax – 12a – 2bx + 6b
7.
x3 – 8
8.
6y2 – 48y + 72
9.
10a2 – 5ab – 15b2
10
a4 – 16
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Answers to Practice Exercises
1.
5(x + 1)(x – 1)
2.
(x + 5)(x – 2)
3.
(x + 2)(2x – 1)
4.
x(x + 2)2
5.
2(5 – x)(5 + x)
6.
2(2a – b)(x – 3)
7.
(x – 2)(x2 + 2x + 4)
8.
6(y – 2)(y – 6)
9.
5(2a – 3b)(a + b)
10.
(a2 + 4)(a + 2)(a – 2)
11.
(x – 2)(x + 1)(x – 1)
12.
-3x2y(5x – 4)(3x + 2)
13.
b2(a + 3)2
14.
2a(3a + 2)(9a2 – 6a + 4)
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