1. Factoring Polynomials
MPM 2D
Factoring Polynomials

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

3. 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.
Notice the sign!
2xy + 3y – 4x – 6
= (2xy + 3y) – (4x + 6)
Group terms.
= y (2x + 3) – 2(2x + 3)
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).
= (a – b) (2a – 3c)
Factor.
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Examples: Factor

4. Factoring these trinomials is based on reversing the FOIL process.
To factor a simple 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).
Factoring these trinomials is based on reversing the FOIL process.
Example: Factor x2 + 3x + 2.
Express the trinomial as a product of two binomials with leading term x and unknown constant terms a and b.
x2 + 3x + 2
= (x + a)(x + b) F O I L
= x2
+ bx
+ ax
+ ba
Apply FOIL to multiply the binomials.
= x2 + (b + a) x + ba
Since ab = 2 and a + b = 3, it follows that a = 1 and b = 2.
= x2 + (1 + 2) x + 1 · 2
(Product-sum method)
Therefore, x2 + 3x + 2 = (x + 1)(x + 2).
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Factor x2 + bx + c

5. 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:
= x2 – 8x + 15.
(x – 3)(x – 5) = x2 – 5x – 3x + 15
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Example: Factor

6. two positive factors of 3613 36
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
Sum
Positive Factors of 36
whose sum is 13.
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

7. 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
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

8. Factoring Polynomials of the Form ax2 + bx + c
Factoring complex trinomials of the form ax2 + bx + c, (a  1) can be done by decomposition or cross-check method.
Example: Factor 3x2 + 8x + 4.
3  4 = 12
Decomposition Method
2. We need to find factors of 12
1, 12
2, 6
3, 4
1. Find the product of first and last terms
is 8
whose sum
3. Rewrite the middle term decomposed into the two numbers
3x2 + 2x + 6x + 4
= (3x2 + 2x) + (6x + 4)
4. Factor by grouping in pairs
= x(3x + 2) + 2(3x + 2)
= (3x + 2) (x + 2)
3x2 + 8x + 4 = (3x + 2) (x + 2)
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Factoring Polynomials of the Form ax2 + bx + c

9. Example: Factor 4x2 + 8x – 5. 4  5 = -20 1, 20 -2, 10 4, 5
We need to find factors of 20
1, 20
-2, 10
4, 5
is 8
whose difference
Rewrite the middle term decomposed into the two numbers
4x2 – 2x + 10x – 5
= (4x2 – 2x) + (10x – 5)
Factor by grouping in pairs
= 2x(2x – 1) + 5(2x – 1)
= (2x – 1) (2x + 5)
4x2 + 8x – 5 = (2x –1)(2x – 5)
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Example: Factor