§ 5.4 Factoring Trinomials. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 5.4 Factoring Trinomials In section 5.3, we factored certain polynomials

§ 5.4

Factoring Trinomials

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 5.4


In section 5.3, we factored certain polynomials having four terms using the method of grouping.

Now, we will use trial and error and a problem solving process to factor trinomials.

The polynomials we will begin with will have leading coefficients of one. We will begin by trying to factor thesetrinomials into a product of two binomials. Polynomialsthat cannot be factored over a given number set are said to be prime.



A Strategy for Factoring T1) Enter x as the first term of each factor.

2) List pairs of factors of the constant c.

3) Try various combinations of these factors. Select the combination in which the sum of the Outside and Inside products is equal to bx.

4) Check your work by multiplying the factors using the FOIL method. You should obtain the original trinomial.

If none of the possible combinations yield an Outside product and an Inside product who sum is equal to bx, the trinomial cannot be factored using integers and is called prime over the set of integers.

cbxx 2

cbxxxx 2

cbxxxx 2

I

OSum of O + I



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.xx 24112

xxxx 24112

Factor:

1) Enter x as the first term of each factor.

To find the second term of each factor, we must find two integers whose product is 24 and whose sum is 11.

2) List pairs of factors of the constant, 24.

Some Factors of 24 24, 1 -24, -1 12, 2 -12, -2 8, 3 -8, -3 6, 4 -6, -4



Possible Factorizations of Sum of Outside and Inside Products (Should Equal 11x)

(x + 24)(x + 1) x + 24x = 25x

(x - 24)(x - 1) -x - 24x = -25x

(x + 12)(x + 2) 2x + 12x = 14x

(x - 12)(x - 2) -2x - 12x = -14x

(x + 8)(x + 3) 3x + 8x = 11x

(x - 8)(x - 3) -3x - 8x = -11x

(x + 6)(x + 4) 4x + 6x = 10x

(x - 6)(x - 4) -4x - 6x = 10x

24112 xx

This is the required middle term.

3) Try various combinations of these factors. The correct factorization of is the one in which the sum of the Outside and Inside products is equal to 11x. Here is a list of the possible factorizations.

CONTINUECONTINUEDD

24112 xx



. 3824112 xxxxThus,

Check this result by multiplying the right side using the FOIL method. You should obtain the original trinomial. Because of the commutative property, the factorization can also be expressed as

CONTINUECONTINUEDD

. 8324112 xxxx



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.aa 1452

aaaa 1452

Factor:

1) Enter a as the first term of each factor.

To find the second term of each factor, we must find two integers whose product is -14 and whose sum is 5.

2) & 3) List pairs of factors of the constant, -14, and try various combinations of these factors. Because the desired sum, 5, is positive, the positive factor of -14 must be farther from 0 than the negative factor is. Thus, we will only list pairs of factors of -14 in which the positive factor has the larger absolute value.



. 271452 aaaa

This is the desired sum.

Thus,

CONTINUECONTINUEDD

Some Factors of -14 14, -1 7, -2

Sum of Factors 13 5



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

y.yy 72124 23

yyy 72124 23 1834 2 yyy

Factor:

The GCF of the three terms of the polynomial is 4y. Therefore, we begin by factoring out 4y. Then we factor the remaining trinomial.

Factor out the GCFBegin factoring . Find two integers whose product is -18 and whose sum is 3.

yyy4

634 yyy

. 63472124 23 yyyyyy

1832 yy

The integers are -3 and 6.

Thus,



:2a . 2 cbxaxxx

A Strategy for Factoring TAssume, for the moment, that there is no greatest common factor.

1) Find two First terms whose product is

2) Find two Last terms whose product is c:

3) By trial and error, perform steps 1 and 2 until the sum of the Outside product and Inside product is bx:

cbxax 2

. 2 cbxaxxx

. 2 cbxaxxx

I

O

Sum of O + I

If no such combinations exist, the polynomial is prime.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.xx 15196 2 Factor:

1) Find two First terms whose product is There is more than one choice for our two First terms. Those choices are cataloged below.

.x26

xxxx 615196 2

2) Find two Last terms whose product is 15. There is more than one choice for our two Last terms. Those choices are cataloged below.

15115196 2 xx

xxxx 3215196 2

?

?

?

?

3515196 2 xx



3) Try various combinations of these factors. The correct factorization of is the one in which the sum of the Outside and Inside products is equal to 19x. Here is a list of some of the possible factorizations.

15196 2 xx

CONTINUECONTINUEDD

Possible Factorizations of Sum of Outside & Inside Products (Should Equal 19x)

(6x + 1)(x + 15) 90x + x = 91x

(x + 1)(6x + 15) 15x + 6x = 21x

(3x + 3)(2x + 5) 15x + 6x = 21x

(2x + 3)(3x + 5) 10x + 9x = 19x

(6x + 3)(x + 5) 30x + 3x = 33x

(x + 3)(6x + 5) 5x + 18x = 23x

15196 2 xx




Therefore, the factorization of is:15196 2 xx

CONTINUECONTINUEDD

(2x + 3)(3x + 5) .

Determine which possible factorizations were not represented in the chart on the preceding page.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.yyy 345 31710 Factor:

1) Find two First terms whose product is

3y

yyyy 1031710 2

The GCF of the three terms of the polynomial is . We begin by factoring out .

3y

3171031710 23345 yyyyyy

.10 2y

yyyy 2531710 2 ?

?



2) Find two Last terms whose product is 3. The only possible factorization is (-1)(-3) since the sum of the Outside and Inside products must be -17y, having a negative coefficient.

CONTINUECONTINUEDD

3) Try various combinations of these factors. The correct factorization of is the one in which the sum of the Outside and Inside products is equal to -17y. A list of the possible factorizations can be found on the next page.

31710 2 yy



CONTINUECONTINUEDD Possible Factorizations of Sum of Outside & Inside

Products (Should Equal -17y)

(10y - 1)(y - 3) -30y - y = -31y

(y - 1)(10y - 3) -3y - 10y = -13y

(2y - 1)(5y - 3) -6y - 5y = -11y

(5y - 1)(2y - 3) -15y - 2y = -17yThis is the required middle term.

31710 2 yy

The factorization of is (5y - 1)(2y - 3). Now we include the GCF in the complete factorization of the given polynomial. Thus,

31710 2 yy

. 32153171031710 323345 yyyyyyyyy



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.yxyx 22 81012 Factor:

1) Find two First terms whose product is .12 2x

?

? yxyxyxyx ??1281012 22

yxyxyxyx ?2?681012 22

yxyxyxyx ?3?481012 22 ?

The question marks inside the parentheses indicate that we are looking for the coefficients of y in each factor.



?

? yxyxyxyx ?8?81012 22

yxyxyxyx ?8?81012 22

yxyxyxyx 2?4?81012 22 ?

The question marks inside the parentheses indicate that we are looking for the coefficients of x in each factor.

CONTINUECONTINUEDD

2) Find two Last terms whose product is -8. The possible factorizations are as follows.

yxyxyxyx 2?4?81012 22 ?



CONTINUECONTINUEDD 3) Try various combinations of these factors. The correct

factorization of is the one in which the sum of the Outside and Inside products is equal to 10xy. A list of the possible factorizations can be found below and on the next page.

22 81012 yxyx

Possible Factorizations of Sum of Outside & Inside Products (Should Equal 10xy)

(12x + 8y)(x - y) -12xy + 8xy = -4xy

(x + 8y)(12x - y) -xy + 96xy = 95xy

(12x + 4y)(x - 2y) -24xy + 4xy = -20xy

(x + 4y)(12x - 2y) -2xy + 48xy = 46xy

(6x + 8y)(2x - y) -6xy + 16xy = 10xy

(2x + 8y)(6x - y) -2xy + 48xy = 46xy

22 81012 yxyx



CONTINUECONTINUEDD

Only half of the possible factorizations were checked. For example, (12x – 8y)(x + y) was not checked. It was not necessary since the corresponding factorization (12x + 8y)(x - y) was checked and the absolute value of the coefficient, -4, was not 10.

Possible Factorizations of Sum of Outside & Inside Products (Should Equal 10xy)

(6x + 4y)(2x - 2y) -12xy + 8xy = -4xy

(2x + 4y)(6x - 2y) -4xy + 24xy = 20xy

(4x + 8y)(3x - y) -4xy + 24xy = 20xy

(3x + 8y)(4x - y) -3xy + 32xy = 29xy

(4x + 4y)(3x - 2y) -8xy + 12xy = 4xy

(3x + 4y)(4x - 2y) -6xy + 16xy = 10xy

22 81012 yxyx




CONTINUECONTINUEDD

NOTE: There was another factorization that resulted in the desired middle term, 10xy. That factorization would have worked just as well as the one selected above. That factorization was (6x + 8y)(2x - y) = (2)(3x +4y)(2x – y). The answer we got above is: (3x + 4y)(4x – 2y) = (3x+4y)(2x – y)(2). As you can see, these two answers are equivalent. This illustrates why the greatest common should always be factored out first. Doing this simplifies the factoring process for you.

Thus, . 244381012 22 yxyxyxyx



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.xx 15132 36 Factor:

Notice that the exponent on is half that of the exponent on We will let t equal the variable to the power that is half of 6. Thus, let

3x

.3xt

Therefore,

.6x

15132 323 xx

15132 2 tt

532 tt

532 33 xx . 53215132 3336 xxxx

This is the given polynomial, with written as

6x .

23x

Let Rewrite the trinomial interms of t.

.3xt

Factor the trinomial.

Now substitute for t. 3x



Factoring Using Grouping T1) Multiply the leading coefficient, a, and the constant, c.

2) Find the factors of ac whose sum is b.

3) Rewrite the middle term, bx, as a sum or difference using the factors from step 2.

4) Factor by grouping.

cbxax 2 1a



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.aa 122 Factor by grouping:

The trinomial is of the form

1) Multiply the leading coefficient, a, and the constant, c. Using a = 1 and c = -12.

12121 ac

. 2 cbxax

122 aa

a = 1 b = 1 c = -12



2) Find the factors of ac whose sum is b. We want the factors of -12 whose sum is b, or 1. The factors of -12 whose sum is 1 are 4 and -3.

123412 22 aaaaa

CONTINUECONTINUEDD

3) Rewrite the middle term, a, as a sum or difference using the factors from step 2: 4 and -3.



12342 aaa

CONTINUECONTINUEDD

4) Factor by grouping.

434 aaa

34 aa

Thus, . 34122 aaaa

Group terms

Factor from each group

Factor out a + 4, the common binomial factor

Documents

§ 5.4 Factoring Trinomials. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 5.4 Factoring Trinomials In section 5.3, we factored certain polynomials