Factoring Polynomials

Digital Lesson

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The simplest method of factoring a polynomial is to factor out the greatest common factor (GCF) of each term.

Example: Factor 18x3 + 60x.

GCF = 6x18x3 + 60x = 6x (3x2) + 6x (10)

18x3 = 2 · 3 · 3 · x · x · x

Apply the distributive law

to factor the polynomial.

6x (3x2 + 10) = 6x (3x2) + 6x (10) = 18x3 + 60x

Check the answer by multiplication.

Factor each term.

Find the GCF.

60x = 2 · 2 · 3 · 5 · x

= 6x (3x2 + 10)

= (2 · 3 · x) · 3 · x · x

= (2 · 3 · x) · 2 · 5

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Example: Factor 4x2 – 12x + 20.

GCF = 4.

4(x2 – 3x + 5) = 4x2 – 12x + 20Check 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) = (x + 1) (5 – y)

(x + 1) (5 – y) = 5(x + 1) – y(x + 1)Check.

= 4(x2 – 3x + 5)

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A difference of two squares can be factored using the formula

Example: Factor x2 – 9y2.

= (x)2 – (3y)2

= (x + 3y)(x – 3y)

Write terms as perfect squares.

Use the formula.

The same method can be used to factor any expression which can be written as a difference of squares.

Example: Factor (x + 1)2 – 25y 4.

= (x + 1)2 – (5y2)2

= [(x + 1) + (5y2)][(x + 1) – (5y2)]

= (x + 1 + 5y2)(x + 1 – 5y2)

a2 – b2 = (a + b)(a – b).

x2 – 9y2

(x + 1)2 – 25y 4

D.O.T.S.

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2. Factor 2a2 + 3bc – 2ab – 3ac.

Some polynomials can be factored by grouping terms to produce a common binomial factor.

= 2a2 – 2ab + 3bc – 3ac

= y (2x + 3) – 2(2x + 3)

= (2a2 – 2ab) + (3bc – 3ac)

= 2a(a – b) + 3c(b – a)

= (2xy + 3y) – (4x + 6) Group terms.

Examples: 1. Factor 2xy + 3y – 4x – 6.

Factor each pair of terms.

= (2x + 3) ( y – 2) Factor out the common binomial.

Rearrange terms.

Group terms.

Factor.

= 2a(a – b) – 3c(a – b) b – a = – (a – b).

= (a – b) (2a – 3c) Factor.

Examples: Factor

2xy + 3y – 4x – 6

2a2 + 3bc – 2ab – 3ac

Notice the sign!

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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).

Example: Factor x2 + 3x + 2. = (x + a)(x + b)

Express the trinomial as a product of two binomials with leading term x and unknown constant terms a and b.

= x2

FApply 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

Therefore, x2 + 3x + 2 = (x + 1)(x + 2).

O I L + bx + ax + ba

x2 + 3x + 2

(Product-sum method)

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Example: Factor x2 – 8x + 15.

= (x + a)(x + b)

(x – 3)(x – 5) = x2 – 5x – 3x + 15

x2 – 8x + 15 = (x – 3)(x – 5).

Therefore a + b = -8

Check:

= x2 + (a + b)x + ab

It follows that both a and b are negative.

= x2 – 8x + 15.

SumNegative Factors of 15

-3, - 5 - 8

- 1, - 15 -15

and ab = 15.

x2 – 8x + 15

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Example: Factor x2 + 13x + 36.

= (x + a)(x + b)

Check: (x + 4)(x + 9)

Therefore a and b are:

x2 + 13x + 36

= x2 + 9x + 4x + 36 = x2 + 13x + 36.

= (x + 4)(x + 9)

= x2 + (a + b) x + ab

SumPositive Factors of 36

1, 36 37

153, 12

4, 9 13

6, 6 12

2, 18 20

x2 + 13x + 36

two positive factors of 36

whose sum is 13.

36 13

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Example: Factor 4x3 – 40x2 + 100x.

A polynomial is factored completely when it is written as a product of factors that can not be factored further.

The GCF is 4x.

= 4x(x2 – 10x + 25) Use distributive property to factor out the GCF.

= 4x(x – 5)(x – 5) Factor the trinomial.

4x(x – 5)(x – 5) = 4x(x2 – 5x – 5x + 25)

= 4x(x2 – 10x + 25)

= 4x3 – 40x2 + 100x

4x3 – 40x2 + 100x

Check:

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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.

1. Find the product of

first and last terms

3 4 =

12Decomposition Method

2. We need to find factors of 12whose sum is 8

1, 122, 63, 4

3. Rewrite the middle

term decomposed

into the two numbers

3x2 + 2x + 6x + 4

= x(3x + 2) + 2(3x + 2)

= (3x2 + 2x) + (6x + 4)4. Factor by grouping

in pairs = (3x + 2) (x + 2)

3x2 + 8x + 4 = (3x + 2) (x + 2)

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Example: Factor 4x2 + 8x – 5.

4x2 + 8x – 5 = (2x –1)(2x – 5)

4 5 =

20 We need to find factors of 20

whose difference is 81, 202, 104, 5

Rewrite the middle term

decomposed into the

two numbers

4x2 – 2x + 10x – 5

= 2x(2x – 1) + 5(2x – 1)

= (4x2 – 2x) + (10x – 5)

= (2x – 1) (2x + 5)

Factor by grouping

in pairs