The Greatest Common Factor; Factoring by Grouping Find the greatest common factor of a list of terms. Factor out the greatest common factor. Factor by.

The Greatest Common Factor; Factoring by GroupingFind the greatest common factor of a list of terms.

Factor out the greatest common factor.

Factor by grouping.

Factor trinomials with a coefficient of 1 for the second-degree term.

Factor trinomials with a coefficient greater than 1 for the second degree term.

Factor such trinomials after factoring out the greatest common factor.

Factor difference of squares

Factor perfect square trinomial

6.1

2

3

1

4

5

6

7

8

other factored forms of 12 are

− 6(−2), 3 · 4, −3(−4), 12 · 1, and −12(−1).

The Greatest Common Factor: Factoring by Grouping

To factor means “to write a quantity as a product.” That is, factoring is the opposite of multiplying. For example,

Multiplying Factoring

6 · 2 = 12 12 = 6 · 2

Factors FactorsProduct Product

More than two factors may be used, so another factored form of 12 is2 · 2 · 3. The positive integer factors of 12 are

1, 2, 3, 4, 6, 12.

Find the greatest common factor of a list of terms.

An integer that is a factor of two or more integers is called a common factor of those integers. For example, 6 is a common factor of 18 and 24. Other common factors of 18 and 24 are 1, 2, and 3. The greatest common factor (GCF) of a list of integers is the largest common factor of those integers. Thus, 6 is the greatest common factor of 18 and 24.

Recall that a prime number has only itself and 1 as factors. Factoring numbers into prime factors is the first step in finding the greatest common factor of a list of numbers.

Factors of a number are also divisors of the number. The greatest common factor is actually the same as the greatest common divisor. The are many rules for deciding what numbers to divide into a given number. Here are some especially useful divisibility rules for small numbers.

Find the greatest common factor of a list of terms. (cont’d)

Finding the Greatest Common Factor (GCF)

Step 1: Factor. Write each number in prime factored form.

Step 2: List common factors. List each prime number or each variable that is a factor of every term in the list. (If a prime does not appear in one of the prime factored forms, it cannot appear in the greatest common factor.)

Step 3: Choose least exponents. Use as exponents on the common prime factors the least exponent from the prime factored forms.

Step 4: Multiply. Multiply the primes from Step 3. If there are no primes left after Step 3, the greatest common factor is 1.

Find the greatest common factor of a list of terms. (cont’d)

Find the greatest common factor for each list of numbers.

50, 75

12, 18, 26, 32

22, 23, 24

Solution:

50 52 5 75 53 5

12 22 3

24 2 2 2 3

26 12 3

2 2 223 2 2 18 32 3

GCF = 25

22 2 11

23 1 23

GCF = 2

GCF = 1

EXAMPLE 1 Finding the Greatest Common Factor for Numbers

Find the greatest common factor of a list of terms.

The GCF can also be found for a list of variable terms. The exponent on a variable in the GCF is the least exponent that appears in all the common factors.

Find the greatest common factor of a list of terms. (cont’d)

Solution:

Find the greatest common factor for each list of terms.

9916 1 2 2 22 rr 15 1510 1 52r r 9GCF = 2r

9 15 1216 , 10 , 8r r r

4 5 3 6 9 2, , s t s t s t

12 128 2 22r r

3 2GCF = s t

4 5 4 5s t s t 33 6 6ss t t

9 9 22s t s t

EXAMPLE 2 Finding the Greatest Common Factor for Variable Terms

Writing a polynomial (a sum) in factored form as a product is called

factoring. For example, the polynomial

3m + 12

has two terms: 3m and 12. The GCF of these terms is 3. We can write

3m + 12 so that each term is a product of 3 as one factor.

The polynomial 3m + 12 is not in factored form when written as 3 · m + 3 · 4.The terms are factored, but the polynomial is not. The factored form of 3m +12 is the product 3(m + 4).

3m + 12 = 3 · m + 3 · 4

= 3(m + 4)

The factored form of 3m + 12 is 3(m + 4). This process is called factoring out the greatest common factor.

Distributive property

Factor out the greatest common factor.

GCF = 3

Solution:

245 6 5 2tt t

Write in factored form by factoring out the greatest common factor.

0 21 1r r

22 26x x

52 24 44 2 3pp p q qq 5 2 6 3 4 78 16 12p q p q p q

4 26 12x x

6 5 430 25 10t t t

12 10r r

Be sure to include the 1 in a problem like r12 + r10. Always check that the factored form can be multiplied out to give the original polynomial.

EXAMPLE 3 Factoring Out the Greatest Common Factor

Solution:

6p rq

Write in factored form by factoring out the greatest common factor.

4 3 4 3y y y 4 43y y

6 p q r p q

EXAMPLE 4 Factoring Out the Greatest Common Factor

Factor by grouping.

When a polynomial has four terms, common factors can sometimes be used to factor by grouping.

Factoring a Polynomial with Four Terms by Grouping

Step 1: Group terms. Collect the terms into two groups so that each group has a common factor.

Step 2: Factor within groups. Factor out the greatest common factor from each group.

Step 3: Factor the entire polynomial. Factor out a common binomial factor from the results of Step 2.

Step 4: If necessary, rearrange terms. If Step 2 does not result in a common binomial factor, try a different grouping.

Factor by grouping.

5 2 10pq q p Solution:

5 52p pq 5 2p q

2 3 21 3x xy 2 3 1x y

2 22 3a a ab

2 3 2 3xy y x

22 4 3 6a a ab b

3 23 5 15x x x 2 2 3a a b

2 3 35x xx 23 5x x

EXAMPLE 5 Factoring by Grouping

Factor by grouping.

26 20 15 8y w y yw

Solution:

3 2 5 4 2 5y y w y

26 15 20 8y y w yw

2 5 3 4y y w

9 12 3 4mn m n 3 3 4 1 3 4m n n

3 1 3 4m n

9 4 12 3mn m n

EXAMPLE 6 Rearranging Terms before Factoring by Grouping

Factoring Trinomials

Using the FOIL method, we see that the product of the binomial k − 3 and k +1 is

(k − 3)(k + 1) = k2 − 2k − 3. Multiplying

Suppose instead that we are given the polynomial k2 − 2k − 3 and want to rewrite it as the product (k − 3)(k + 1). That is,

k2 − 2k − 3 = (k − 3)(k + 1). Factoring

Recall that this process is called factoring the polynomial. Factoring reverses or “undoes” multiplying.

Factor trinomials with a coefficient of 1 for the second-degree term.

When factoring polynomials with integer coefficients, we use only integers in the factors. For example, we can factor x2 + 5x + 6 by finding integers m and n such that

x2 + 5x + 6 = (x + m)(x + n).

Comparing this result with x2 + 5x + 6 shows that we must find integers m and n having a sum of 5 and a product of 6.

2 25 6 .x nx xm mnx

Sum of m and n is 5.

Product of m and n is 6.

2 .m n n m mx x x x nx 2 .n m mnx x

To find these integers m and n, we first use FOIL to multiply the two binomials on the right side of the equation:

Since many pairs of integers have a sum of 5, it is best to begin by listing those pairs of integers whose product is 6. Both 5 and 6 are positive, so consider only pairs in which both integers are positive.

Both pairs have a product of 6, but only the pair 2 and 3 has a sum of 5. So 2 and 3 are the required integers, and

x2 + 5x + 6 = (x + 2)(x + 3).

Check by using the FOIL method to multiply the binomials. Make sure that the sum of the outer and inner products produces the correct middle term.

Factor trinomials with a coefficient of 1 for the second-degree term. (cont’d)

Factor y2+ 12y + 20.

Solution:

10 2y y

Factors of 20 Sums of Factors

1, 20 1 + 20 = 21

2, 10 2 + 10 = 12

4, 5 4 + 5 = 9

You can check your factoring by graphing both the unfactored and factored forms of polynomials on your graphing calculators.

EXAMPLE 1 Factoring a Trinomial with All Positive Terms

Factor y2 − 10y + 24.

Solution:

6 4y y

Factors of 24 Sums of Factors

− 1 , −24 −1 + (−24) = −25

−2 , −12 −2 + (−12) = −14

−3 , −8 −3 + (−8) = −11

−4 , −6 −4 + (−6) = −10

EXAMPLE 2 Factoring a Trinomial with a Negative Middle Term

Factor z2 + z − 30.

Solution:

6 5z z

Factors of − 30 Sums of Factors

− 1 , 30 −1 + (30) = 29

1 , − 30 1 + (−30) = −29

5 , − 6 5 + (− 6) = −1

−5 , 6 −5 + (6) = 1

EXAMPLE 3 Factoring a Trinomial with a Negative Last (Constant) Term

Factor a2 − 9a − 22.

Solution:

11 2a a

Factors of −22 Sums of Factors

−1 , 22 −1 + 22 = 21

1, −22 1 + (−22) = −21

−2 , 11 −2 + 11 = 9

2 , −11 2 + (−11) = −9

EXAMPLE 4 Factoring a Trinomial with Two Negative Terms

Some trinomials cannot be factored by using only integers. We call such trinomials prime polynomials.

Factor trinomials with a coefficient of 1 for the second-degree term. (cont’d)

Summarize the signs of the binomials when factoring a trinomial whose leading coefficient is positive.

1. If the last term of the trinomial is positive, both binomials will have the same “middle” sign as the second term.

2. If the last term of the trinomial is negative, the binomials will have one plus and one minus “middle” sign.

Factor if possible.

Factors of 14 Sums of Factors

−1 , −14 −1 + (−14) =

−15

−2 , −7 −2 + (−7) = −9

Solution:

2 8 14m m

2 2y y

Prime

Prime

Factors of 2 Sums of Factors

1, 2 1 + 2 = 3

EXAMPLE 5 Deciding Whether Polynomials Are Prime

Guidelines for Factoring x2 + bx + c

Find two integers whose product is c and whose sum is b.

1. Both integers must be positive if b and c are positive.

2. Both integers must be negative if c is positive and b is negative.

3. One integer must be positive and one must be negative if c is negative.

Factor trinomials with a coefficient of 1 for the second-degree term. (cont’d)

Factor trinomials with a coefficient greater than 1 for the second-degree term. (Slide and Divide)

Factor : 10x2 – 21x -10

Step 4: Multiply each binomial by the denominator of its constant term.

(2x-5)(5x+2) Check by multiplying

Step 1: Multiply second degree coefficient and constant term. Remove second degree term and replace constant term with new constant term.

x2 – 21x – 100

Step 2: Factor as if second degree coefficient is 1.(x–25)(x+4) -25 + 4 = -21; -25 x 4 = -100

Step 3: Divide each binomials constant term by original second degree coefficient. Reduce fractions if possible.

(x-25/10)(x+4/10) (x-5/2)(x+2/5)

Factor r2 − 6rs + 8s2.

Solution:

4 2r s r s

Factors of 8s2 Sums of Factors

−1s , −8s − 1s + (−8s) = −9s

−2s , −4s −2s + (−4s) = −6s

EXAMPLE 6 Factoring a Trinomial with Two Variables

Factor 6m2 + 11mn – 10n2.

3 2 2 5m n m n 4mn

15mn11mn

Solution:

EXAMPLE 6 Factoring a Trinomial with Two Variables

Factor 3x4 − 15x3 + 18x2.

2 23 5 6x x x

23 3 2x x x

Solution:

When factoring, always look for a common factor first. Remember to include the common factor as part of the answer. As a check, multiplying out the factored form should always give the original polynomial.

EXAMPLE 7 Factoring a Trinomial with a Common Factor

Factor such trinomials after factoring out the greatest common factor.If a trinomial has a common factor, first factor it out.

Factor.

Solution:

4 3 228 58 30x x x 3 2 224 32 6x x y xy

22 14 29 152 xx x

22 7 3 2 5x x x

2 212 16 32 x yx xy

2 6 2 3x x y x y

6x35x

29x

2xy18xy

16xy

EXAMPLE 7 Factoring Trinomials with Common Factors

Factor a difference of squares.

The formula for the product of the sum and difference of the same two terms is

Factoring a Difference of Squares

2 2.x y x y x y

2 2x y x y x y

The following conditions must be true for a binomial to be a difference of squares:

1. Both terms of the binomial must be squares, such as

x2, 9y2, 25, 1, m4.2. The second terms of the binomials must have different signs (one

positive and one negative).

2 2 216 4 4 4 .m m m m For example,

Factor each binomial if possible.

Solution:

2 81t

2 2r s2 10y 2 36q

9 9t t

r s r s

prime

prime

After any common factor is removed, a sum of squares cannot be factored.

EXAMPLE 1 Factoring Differences of Squares

Factor each difference of squares.

Solution:

249 25x

2 264 81a b

7 5 7 5x x

8 9 8 9a b a b

You should always check a factored form by multiplying.

EXAMPLE 2 Factoring Differences of Squares

Factor completely.

Solution:

250 32r

4 100z

4 81z

22 25 16r

2 210 10z z

2 29 9z z

2 5 4 5 4r r

Factor again when any of the factors is a difference of squares as in the last problem.

Check by multiplying.

2 9 3 3z z z

EXAMPLE 3 Factoring More Complex Differences of Squares

The expressions 144, 4x2, and 81m6 are called perfect squares because

A perfect square trinomial is a trinomial that is the square of a binomial. A necessary condition for a trinomial to be a perfect square is that two of its terms be perfect squares.

Even if two of the terms are perfect squares, the trinomial may not be a perfect square trinomial.

Factor a perfect square trinomial.

Factoring Perfect Square Trinomials

22 22x xy y x y

22 22x xy y x y

2144 ,12 224 2 ,x x 26 381 9 .m mand

Solution:

210k

Factor k2 + 20k + 100.

2 20 100k k

Check :

2 10 20k k

EXAMPLE 4 Factoring a Perfect Square Trinomial

Factor each trinomial.

Solution:

2 24 144x x

225 30 9x x

236 20 25a a

212x

25 3x

prime

3 218 84 98x x x 22 9 42 49x x x 22 3 7x x

EXAMPLE 5 Factoring Perfect Square Trinomials

1. The sign of the second term in the squared binomial is always the same as the sign of the middle term in the trinomial.

Factoring Perfect Square Trinomials

3. Perfect square trinomials can also be factored by using grouping or the FOIL method, although using the method of this section is often easier.

2. The first and last terms of a perfect square trinomial must be positive, because they are squares. For example, the polynomial x2 – 2x – 1 cannot be a perfect square, because the last term is negative.

Factor a difference of cubes.

The polynomial x3 − y3 is not equivalent to (x − y )3,

whereas

3x y x y x y x y

2 22x y x xy y

3 3 2 2x y x y x xy y

Factoring a Difference of Cubes

3 3 2 2x y x y x xy y same sign

positive

opposite sign

This pattern for factoring a difference of cubes should be memorized.

Factor each polynomial.

Solution:

3 216x

327 8x 35 5x

26 6 36x x x

23 2 9 6 4x x x

3 664 125x y 2 2 2 44 5 16 20 25 x y x xy y

35 1x 25 1 1x x x

A common error in factoring a difference of cubes, such as

x3 − y3 = (x − y)(x2 + xy + y2), is to try to factor x2 + xy + y2. It is easy to confuse

this factor with the perfect square trinomial x2 + 2xy + y2. But because there is

no 2, it is unusual to be able to further factor an expression of the form x2 +

xy +y2.

EXAMPLE 6 Factoring Differences of Cubes

Factor a sum of cubes.

A sum of squares, such as m2 + 25, cannot be factored by using real numbers, but a sum of cubes can.

Note the similarities in the procedures for factoring a sum of cubes and a difference of cubes.

1. Both are the product of a binomial and a trinomial.

2. The binomial factor is found by remembering the “cube root, same sign, cube root.”

3. The trinomial factor is found by considering the binomial factor and remembering, “square first term, opposite of the product, square last term.”

Factoring a Sum of Cubes

3 3 2 2x y x y x xy y

same sign

positive

opposite sign

Methods of factoring discussed in this section.

Factor each polynomial.

Solution:

3 64p

3 327 64x y

6 3512a b

24 4 16p p p

2 23 4 9 12 16x y x xy y

2 4 2 28 64 8a b a a b b

EXAMPLE 7 Factoring Sums of Cubes