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Factoring Trinomials with Leading Coefficient 1

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken great care in developing the student's understanding of the factorization process. The technique is consistently illustrated by displaying an empty set of parentheses and describing the thought process used to discover the terms that are to be placed inside the parentheses. The factoring scheme for special products is presented with both verbal and symbolic descriptions, since not all students can interpret symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with leading coefficients different from 1. Objectives of this module: be able to factor trinomials with leading coefficient 1, become familiar with some factoring hints.

Overview

  • Method
  • Factoring Hints

Method

Let’s consider the product of the two binomials ( x+4 ) ( x 4 ) and ( x+7 ) ( x 7 ) .

The product of two binomials, x plus four and x plus seven, is equal to x squared plus seven x plus four x plus twenty eight, which is simplified to x squared plus eleven x plus twenty eight. The FOIL method is shown by arrows from the first binomial to the second binomial in the product.

Notice that the first term in the resulting trinomial comes from the product of the first terms in the binomials: xx= x 2 xx= x 2 . The last term in the trinomial comes from the product of the last terms in the binomials: 47=28 47=28 . The middle term comes from the addition of the outer and inner products: 7 x+4 x=11 x 7 x 4 x 11 x. Also, notice that the coefficient of the middle term is exactly the sum of the last terms in the binomials: 4+7=11 4 7 11 .

The problem we’re interested in is that given a trinomial, how can we find the factors? When the leading coefficient (the coefficient of the quadratic term) is 1, the observations we made above lead us to the following method of factoring.

Method of Factoring

  1. Write two sets of parentheses: ( )( ) ( )( ) .
  2. Place a binomial into each set of parentheses. The first term of each binomial is a factor of the first term of the trinomial.
  3. Determine the second terms of the binomials by determining the factors of the third term that when added together yield the coefficient of the middle term.

Sample Set A

Factor the following trinomials.

Example 1

x 2 +5x+6 x 2 +5x+6

  1. Write two sets of parentheses: ( )( ) ( )( ) .
  2. Place the factors of x 2 x 2 into the first position of each set of parentheses:

    (x )(x ) (x )(x )

  3. The third term of the trinomial is 6. We seek two numbers whose
    (a) product is 6 and
    (b) sum is 5.
    The required numbers are 3 and 2. Place +3and+2 +3and+2 into the parentheses.

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

    The factorization is complete. We’ll check to be sure.

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

Example 2

y 2 2y24 y 2 2y24

  1. Write two sets of parentheses: ( )( ) ( )( ) .
  2. Place the factors of y 2 y 2 into the first position of each set of parentheses:

    (y )(y ) (y )(y )

  3. The third term of the trinomial is 24 24 . We seek two numbers whose
    (a) product is 24 24 and
    (b) sum is 2 2 .
    The required numbers are 6and4 6and4 . Place 6and+4 6and+4 into the parentheses.

    y 2 2y24=(y6)(y+4) y 2 2y24=(y6)(y+4)

    The factorization is complete. We’ll check to be sure.

    ( y 6)(y+4) = y 2 +4y6y24 = y 2 2y24 ( y 6)(y+4) = y 2 +4y6y24 = y 2 2y24

Notice that the other combinations of the factors of 24 24 (some of which are 2,12;3,8;and4,6 2,12;3,8;and4,6 ) do not work. For example,

(y2)(y+12) = y 2 + 10y 24 (y+3)(y8) = y 2 5y 24 (y4)(y+6) = y 2 + 2y 24 (y2)(y+12) = y 2 + 10y 24 (y+3)(y8) = y 2 5y 24 (y4)(y+6) = y 2 + 2y 24

In all of these equations, the middle terms are incorrect.

Example 3

a 2 11a+30 a 2 11a+30

  1. Write two sets of parentheses: ( )( ) ( )( ) .
  2. Place the factors of a 2 a 2 into the first position of each set of parentheses:

    (a )(a ) (a )(a )

  3. The third term of the trinomial is +30 +30 . We seek two numbers whose
    (a) product is 30 and
    (b) sum is 11 11 .
    The required numbers are 5and6 5and6 . Place 5and6 5and6 into the parentheses.

    a 2 11a+30=(a5)(a6) a 2 11a+30=(a5)(a6)

    The factorization is complete. We’ll check to be sure.

    (a5)(a6) = a 2 6a5a+30 = a 2 11a+30 (a5)(a6) = a 2 6a5a+30 = a 2 11a+30

Example 4

3 x 2 15x42 3 x 2 15x42

Before we begin, let’s recall the most basic rule of factoring: factor out common monomial factors first. Notice that 3 is the greatest common monomial factor of every term. Factor out 3.

3 x 2 15x42=3( x 2 5x14) 3 x 2 15x42=3( x 2 5x14)

Now we can continue.

  1. Write two sets of parentheses: 3( )( ) 3( )( ) .
  2. Place the factors of x 2 x 2 into the first position of each set of parentheses:

    3 (x )(x ) 3 (x )(x )

  3. The third term of the trinomial is 14 14 . We seek two numbers whose
    (a) product is 14 14 and
    (b) sum is 5 5 .
    The required numbers are 7and2 7and2 . Place 7and+2 7and+2 into the parentheses.

    3 x 2 15x42=3(x7)(x+2) 3 x 2 15x42=3(x7)(x+2)

    The factorization is complete. We’ll check to be sure.

    3(x7)(x+2) = 3( x 2 +2x7x14) = 3( x 2 5x14) = 3 x 2 15x42 3(x7)(x+2) = 3( x 2 +2x7x14) = 3( x 2 5x14) = 3 x 2 15x42

Practice Set A

Factor, if possible, the following trinomials.

Exercise 1

k 2 +8 k+15 k 2 8 k 15

Solution

(k+3)(k+5) (k+3)(k+5)

Exercise 2

y 2 +7 y30 y 2 7 y 30

Solution

(y+10)(y3) (y+10)(y3)

Exercise 3

m 2 +10 m+24 m 2 10 m 24

Solution

(m+6)(m+4) (m+6)(m+4)

Exercise 4

m 2 10 m+16 m 2 10 m 16

Solution

(m8)(m2) (m8)(m2)

Factoring Hints

Factoring trinomials may take some practice, but with time and experience, you will be able to factor much more quickly.

There are some clues that are helpful in determining the factors of the third term that when added yield the coefficient of the middle term.

Factoring Hints

Look at the sign of the last term:

  1. If the sign is positive, we know that the two factors must have the same sign, since (+)(+)=(+)and()()=(+) (+)(+)=(+)and()()=(+) . The two factors will have the same sign as the sign of the middle term.
  2. If the sign is negative, we know that two factors must have opposite signs, since (+)()=()and()(+)=() (+)()=()and()(+)=() .

Sample Set B

Example 5

Factor x 2 7x+12 x 2 7x+12 .

  1. Write two sets of parentheses: ( )( ) ( )( ) .
  2. The third term of the trinomial is +12 +12 . The sign is positive, so the two factors of 12 we are looking for must have the same sign. They will have the sign of the middle term. The sign of the middle term is negative, so both factors of 12 are negative. They are 12and1,6and2,or4and3 12and1,6and2,or4and3 . Only the factors 4and3addto7,so4and3 4and3addto7,so4and3 are the proper factors of 12 to be used.

    x 2 7x+12=(x4)(x3) x 2 7x+12=(x4)(x3)

Practice Set B

Factor, if possible, the following trinomials.

Exercise 5

4 k 2 +32 k+28 4 k 2 32 k 28

Solution

4(k+7)(k+1) 4(k+7)(k+1)

Exercise 6

3 y 4 +24 y 3 +36 y 2 3 y 4 24 y 3 36 y 2

Solution

3 y 2 (y+2)(y+6) 3 y 2 (y+2)(y+6)

Exercise 7

x 2 x y6 y 2 x 2 x y 6 y 2

Solution

(x+2y)(x3y) (x+2y)(x3y)

Exercise 8

5 a 5 b10 a 4 b 2 +15 a 3 b 3 5 a 5 b 10 a 4 b 2 15 a 3 b 3

Solution

5 a 3 b(a+3b)(ab) 5 a 3 b(a+3b)(ab)

Exercises

For the following problems, factor the trinomials when possible.

Exercise 9

x 2 +4x+3 x 2 +4x+3

Solution

( x+3 )( x+1 ) ( x+3 )( x+1 )

Exercise 10

x 2 +6x+8 x 2 +6x+8

Exercise 11

x 2 +7x+12 x 2 +7x+12

Solution

( x+3 )( x+4 ) ( x+3 )( x+4 )

Exercise 12

x 2 +6x+5 x 2 +6x+5

Exercise 13

y 2 +8y+12 y 2 +8y+12

Solution

( y+6 )( y+2 ) ( y+6 )( y+2 )

Exercise 14

y 2 5y+6 y 2 5y+6

Exercise 15

y 2 5y+4 y 2 5y+4

Solution

( y4 )( y1 ) ( y4 )( y1 )

Exercise 16

a 2 +a6 a 2 +a6

Exercise 17

a 2 +3a4 a 2 +3a4

Solution

( a+4 )( a1 ) ( a+4 )( a1 )

Exercise 18

x 2 +4x21 x 2 +4x21

Exercise 19

x 2 -4x21 x 2 -4x21

Solution

( x7 )( x+3 ) ( x7 )( x+3 )

Exercise 20

x 2 +7x+12 x 2 +7x+12

Exercise 21

y 2 +10y+16 y 2 +10y+16

Solution

( y+8 )( y+2 ) ( y+8 )( y+2 )

Exercise 22

x 2 +6x-16 x 2 +6x-16

Exercise 23

y 2 -8y+7 y 2 -8y+7

Solution

( y7 )( y1 ) ( y7 )( y1 )

Exercise 24

y 2 5y24 y 2 5y24

Exercise 25

a 2 +a30 a 2 +a30

Solution

( a+6 )( a5 ) ( a+6 )( a5 )

Exercise 26

a 2 3a+2 a 2 3a+2

Exercise 27

a 2 12a+20 a 2 12a+20

Solution

( a10 )( a2 ) ( a10 )( a2 )

Exercise 28

y 2 4y32 y 2 4y32

Exercise 29

x 2 +13x+42 x 2 +13x+42

Solution

( x+6 )( x+7 ) ( x+6 )( x+7 )

Exercise 30

x 2 +2x35 x 2 +2x35

Exercise 31

x 2 +13x+40 x 2 +13x+40

Solution

( x+5 )( x+8 ) ( x+5 )( x+8 )

Exercise 32

y 2 +6y27 y 2 +6y27

Exercise 33

b 2 +15b+56 b 2 +15b+56

Solution

( b+8 )( b+7 ) ( b+8 )( b+7 )

Exercise 34

3 a 2 +24a+36 3 a 2 +24a+36
(Hint: Always search for a common factor.)

Exercise 35

4 x 2 +12x+8 4 x 2 +12x+8

Solution

4( x+2 )( x+1 ) 4( x+2 )( x+1 )

Exercise 36

2 a 2 18a+40 2 a 2 18a+40

Exercise 37

5 y 2 70y+440 5 y 2 70y+440

Solution

5( y 2 14y+88 ) 5( y 2 14y+88 )

Exercise 38

6 x 2 54x+48 6 x 2 54x+48

Exercise 39

x 3 +6 x 2 +8x x 3 +6 x 2 +8x

Solution

x( x+4 )( x+2 ) x( x+4 )( x+2 )

Exercise 40

x 3 8 x 2 +15x x 3 8 x 2 +15x

Exercise 41

x 4 +9 x 3 +14 x 2 x 4 +9 x 3 +14 x 2

Solution

x 2 ( x+7 )( x+2 ) x 2 ( x+7 )( x+2 )

Exercise 42

2 a 3 +12 a 2 +10a 2 a 3 +12 a 2 +10a

Exercise 43

4 a 3 40 a 2 +84a 4 a 3 40 a 2 +84a

Solution

4a( a7 )( a3 ) 4a( a7 )( a3 )

Exercise 44

3x m 2 +33xm+54x 3x m 2 +33xm+54x

Exercise 45

2 y 2 n 2 10 y 2 n48 y 2 2 y 2 n 2 10 y 2 n48 y 2

Solution

2 y 2 ( n8 )( n+3 ) 2 y 2 ( n8 )( n+3 )

Exercise 46

4 x 4 42 x 3 +144 x 2 4 x 4 42 x 3 +144 x 2

Exercise 47

y 5 +13 y 4 +42 y 3 y 5 +13 y 4 +42 y 3

Solution

y 3 ( y+6 )( y+7 ) y 3 ( y+6 )( y+7 )

Exercise 48

4 x 2 a 6 48 x 2 a 5 +252 x 2 a 4 4 x 2 a 6 48 x 2 a 5 +252 x 2 a 4

Exercises for Review

Exercise 49

((Reference)) Factor 6 x y+2 a x3 a y a 2 6x y 2 a x 3 a y a 2 .

Solution

( 2xa )( 3y+a ) ( 2xa )( 3y+a )

Exercise 50

((Reference)) Factor 8 a 2 50 8 a 2 50 .

Exercise 51

((Reference)) Factor 4 x 2 +17 x15 4 x 2 17 x 15 .

Solution

( 4x3 )( x+5 ) ( 4x3 )( x+5 )

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