1. Factoring Polynomials
Digital Lesson
Factoring Polynomials

2. Greatest Common Factor
The simplest method of factoring a polynomial is to factor out the greatest common factor (GCF) of each term.
Example: Factor 18x3 + 60x.
18x3 = 2 · 3 · 3 · x · x · x
Factor each term.
= (2 · 3 · x) · 3 · x · x
60x = 2 · 2 · 3 · 5 · x
= (2 · 3 · x) · 2 · 5
GCF = 6x
Find the GCF.
18x3 + 60x = 6x (3x2) + 6x (10)
Apply the distributive law to factor the polynomial.
= 6x (3x2 + 10)
Check the answer by multiplication.
6x (3x2 + 10) = 6x (3x2) + 6x (10) = 18x3 + 60x
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Greatest Common Factor

3. A common binomial factor can be factored out of certain expressions.
Example: Factor 4x2 – 12x + 20.
4x2 = 2 · 2 · x · x
, 12x = 2 · 2 · 3 · x
, 20 = 2 · 2 · 5
Therefore, GCF = 4.
4x2 – 12x + 20 = 4x2 – 4 · 3x + 4 · 5
= 4(x2 – 3x + 5)
4(x2 – 3x + 5) = 4x2 – 12x + 20
Check the answer.
A common binomial factor can be factored out of certain expressions.
Example: Factor the expression 5(x + 1) – y(x + 1).
5(x + 1) – y(x + 1) = (5 – y)(x + 1)
(5 – y)(x + 1) = 5(x + 1) – y(x + 1)
Check.
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Example: Factor

4. A difference of squares can be factored using the formula
a2 – b2 = (a + b)(a – b).
Example: Factor x2 – 9y2.
x2 – 9y2
= (x)2 – (3y)2
Write terms as perfect squares.
= (x + 3y)(x – 3y)
Use the formula.
The same method can be used to factor any expression which can be written as a difference of squares.
Example: Factor 4(x + 1)2 – 25y 4.
4(x + 1)2 – 25y 4
= (2(x + 1))2 – (5y2)2
= [(2(x + 1)) + (5y2)][(2(x + 1)) – (5y2)]
= (2x y2)(2x + 2 – 5y2)
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Difference of Squares

5. Examples: 1. Factor 2xy + 3y – 4x – 6.
Some polynomials can be factored by grouping terms to produce a common binomial factor.
Examples: 1. Factor 2xy + 3y – 4x – 6.
2xy + 3y – 4x – 6
= (2xy + 3y) – (4x + 6)
Group terms.
= (2x + 3)y – (2x + 3)2
Factor each pair of terms.
= (2x + 3)( y – 2)
Factor out the common binomial.
2. Factor 2a2 + 3bc – 2ab – 3ac.
2a2 + 3bc – 2ab – 3ac
= 2a2 – 2ab + 3bc – 3ac
Rearrange terms.
= (2a2 – 2ab) + (3bc – 3ac)
Group terms.
= 2a(a – b) + 3c(b – a)
Factor.
= 2a(a – b) – 3c(a – b)
b – a = – (a – b).
= (2a – 3c)(a – b)
Factor.
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Examples: Factor

6. Sum or Difference of Cubes
A sum or difference of cubes can be factored using the formulas
a3 + b3 = (a + b)(a2 – ab + b2) and a3 – b3 = (a – b)(a2 + ab + b2).
Examples: 1. Factor x3 + 8y3.
x3 + 8y3
= (x)3 + (2y)3
Write each term as a perfect cube.
= (x + 2y)(x2 – x(2y) + (2y)2)
Use the sum formula.
= (x + 2y)(x2 – 2xy + 4y2)
Simplify.
2. Factor 27a3 – y3z6.
27a3 – y3z6
= (3a)3 – ( yz2)3
Write each term as a perfect cube.
= ((3a) – ( yz2))((3a)2 + (3a)( yz2) +( yz2)2)
Use the difference formula.
= (3a – yz2)(9a2 + 3ayz2 + y2z 4)
Simplify.
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Sum or Difference of Cubes

7. Therefore, x2 + 3x + 2 = (x + 1)(x + 2).
To factor a trinomial of the form x2 + bx + c, express the trinomial as the product of two binomials. For example,
x2 + 10x + 24 = (x + 4)(x + 6).
One method of factoring trinomials is based on reversing the FOIL process.
Express the trinomial as a product of two binomials with leading term x and unknown constant terms a and b.
Example: Factor x2 + 3x + 2.
x2 + 3x + 2
= (x + a)(x + b) F O I L
= x2
+ ax
+ bx
+ ab
Apply FOIL to multiply the binomials.
= x2 + (a + b) x + ab
Since ab = 2 and a + b = 3, it follows that a = 1 and b = 2.
= x2 + (1 + 2) x + 1 · 2
Therefore, x2 + 3x + 2 = (x + 1)(x + 2).
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Factor x2 + bx + c

8. It follows that both a and b are negative.
Example: Factor x2 – 8x + 15.
x2 – 8x + 15
= (x + a)(x + b)
= x2 + (a + b)x + ab
Therefore a + b = – 8
and ab = 15.
It follows that both a and b are negative.
Sum
Negative Factors of 15
– 1, – 15
+15
– 3, – 5
– 8
x2 – 8x + 15 = (x – 3)(x – 5).
Check:
(x – 3)(x – 5) = x2 – 5x – 3x + 15
= x2 – 8x + 15.
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Example: Factor

9. Therefore a and b are two positive factors of 36 whose sum is 13.
Example: Factor x2 + 13x + 36.
x2 + 13x + 36
= (x + a)(x + b)
= x2 + (a + b) x + ab
Therefore a and b are two positive factors of 36 whose sum is 13.
Sum
Positive Factors of 36
1, 36 37
2, 18 20 15
3, 12
4, 9 13
6, 6 12
= (x + 4)(x + 9)
x2 + 13x + 36
Check: (x + 4)(x + 9)
= x2 + 9x + 4x + 36
= x2 + 13x + 36.
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Example: Factor

10. Example: Factor Completely
A polynomial is factored completely when it is written as a product of factors that can not be factored further.
Example: Factor 4x3 – 40x x.
4x3 – 40x x
= 4x(x2) – 4x(10x) + 4x(25)
The GCF is 4x.
= 4x(x2 – 10x + 25)
Use distributive property to factor out the GCF.
= 4x(x – 5)(x – 5)
Factor the trinomial.
Check:
4x(x – 5)(x – 5)
= 4x(x2 – 5x – 5x + 25)
= 4x(x2 – 10x + 25)
= 4x3 – 40x x
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Example: Factor Completely

11. Factoring Polynomials of the Form ax2 + bx + c
Factoring polynomials of the form ax2 + bx + c, (a  1) is usually done by trial and error.
Example: Factor 2x2 + 5x + 3.
2x2 + 5x + 3
= (2x + a)(x + b)
For some a and b.
= 2x2 + (a + 2b)x + ab
Therefore, a + 2b = 5 and ab = 3.
a b a + 2b
– – – 7 5
2x2 + 5x + 3 = (2x + 3)(x + 1)
– – – 5
Check: (2x + 3)(x + 1) = 2x2 + 2x + 3x + 3 = 2x2 + 5x + 3.
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Factoring Polynomials of the Form ax2 + bx + c

12. This polynomial factors as (x + a)(4x + b) or (2x + a)(2x + b).
Example: Factor 4x2 – 12x + 5.
This polynomial factors as (x + a)(4x + b) or (2x + a)(2x + b).
Since ab = + 5, a and b have the same sign.
a = –1, b = – 5 or a = 1 and b = 5.
The middle term –12x equals either (4a + b) x or (2a + 2b) x. Since a and b cannot both be positive, they must both be negative.
Trial Factors
Middle Term
(x –1)(4x – 5)
– 5x – 4x = – 9x
Create trial factorizations with the forms above.
(x –5)(4x – 1)
– x – 20x = – 21x
(2x –1)(2x – 5)
– 10x – 2x = –12x
4x2 – 12x + 5 = (2x –1)(2x – 5)
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Example: Factor

13. Example: Factoring Trinomials
Trinomials which are quadratic in form are factored like quadratic trinomials.
Example: Factor 3x x2 + 9.
3x x2 + 9
= 3u2 + 28u + 9
Let u = x2.
= (3u + 1)(u + 9)
Factor.
= (3x2 + 1)(x2 + 9)
Replace u by x2.
Many trinomials cannot be factored.
Example: Factor x2 + 3x + 5.
Let x2 + 3x + 5 = (x + a)(x + b) = x2 + (a + b) x + ab.
Then a + b = 3 and ab = 5. This is impossible.
The trinomial x2 + 3x + 5 cannot be factored.
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Example: Factoring Trinomials