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6.7: Factoring Trinomials with Leading Coefficient 1

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Method

Let’s consider the product of the two binomials (x+4) and (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=x2. The last term in the trinomial comes from the product of the last terms in the binomials: 47=28. The middle term comes from the addition of the outer and inner products: 7x+4x=11x. Also, notice that the coefficient of the middle term is exactly the sum of the last terms in the binomials: 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 6.7.1

x2+5x+6

1. Write two sets of parentheses: ( ) ( ).

2. Place the factors of x2 into the first position of each set of parentheses:
(x ) ( x )
3. The third term of the trinomial is 6. We seek two numbers whose
(a) product is 6
(b) sum is 5
The required numbers are 3 and 2. Place +3 and +2 into the parentheses.
x2+5x+6=(x+3)(x+2)
The factorization is complete. We'll check to be sure:

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

Example 6.7.2

y22y24

1. Write two sets of parentheses: ( ) ( ).

2. Place the factors of y2 into the first position of each set of parentheses:
(y ) ( y )

3. The third term of the trinomial is 24. We seek two numbers whose
(a) product is 24 and
(b) sum is 2
The required numbers are 6 and 4. Place 6 and +4 into the parentheses.
y22y24=(y6)(y+4)
The factorization is complete. We'll check to be sure.

(y6)(y+4)=y2+4y6y24=y22y24

Notice that the other combinations of 24 (some of which are 2,12;3,8;4,6), do not work. For example,

(y2)(y+12)=y2+10y24(y+3)(y8)=y25y24(y4)(y+6)=y2+2y24

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

Example 6.7.3

a211a+30

1. Write two sets of parentheses: ( ) ( ).

2. Place the factors of a2 into the first position of each set of parentheses:
(a ) ( a )

3. The third term of the trinomial is +30. We seek two numbers whose:
(a) product is 30 and
(b) sum is 11.
The required numbers are 5 and 6. Place 5 and 6 into the parentheses.
a211a+30=(a5)(a6)
The factorization is complete. We'll check to be sure

(a5)(a6)=a26a5a+30=a211a+30

Example 6.7.4

3x215x42

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.

3x215x42=3(x25x14)

Now we can continue.

1. Write two sets of parentheses: 3( ) ( ).

2. Place the factors of x2 into the first position of each set of parentheses:
3(x ) ( x )

3. The third term of the trinomial is 14. We seek two numbers whose
(a) product is 14 and
(b) sum is 5.
The required numbers are 7 and 2. Place 7 and +2 into the parentheses.
3x215x42=3(x7)(x+2)
The factorization is complete. We'll check to be sure.

3(x7)(x+2)=3(x2+2x7x14)=3(x25x14)=3x215x42

Practice Set A

Factor, if possible, the following trinomials.

Practice Problem 6.7.1

k2+8k+15

Answer

(k+3)(k+5)

Practice Problem 6.7.2

y2+7y30

Answer

(y+10)(y3)

Practice Problem 6.7.3

m2+10m+24

Answer

(m+6)(m+4)

Practice Problem 6.7.4

m210m+16

Answer

(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 ()()=(+). 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 ()(+)=().

Sample Set B

Example 6.7.5

Factor x27x+12

1. Write two sets of parentheses: ( ) ( ).

2. The third term of the trinomial is +12. The sign is potitive, 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 12 and 1, 6 and 2, or 4 and 3. Only the factors 4 and 3 add to 7, so 4 and 3 are the proper factors of 12 to be used.

x27x+12=(x4)(x3)

Practice Set B

Factor, if possible, the following trinomials.

Practice Problem 6.7.5

4k2+32k+28

Answer

4(k+7)(k+1)

Practice Problem 6.7.5

3y4+24y3+36y2

Answer

3y2(y+2)(y+6)

Practice Problem 6.7.5

x2xy6y2

Answer

(x+2y)(x3y)

Practice Problem 6.7.5

5a5b10a4b2+15a3b3

Answer

5a3b(a+3b)(ab)

Exercises

For the following problems, factor the trinomials when possible.

Exercise 6.7.1

x2+4x+3

Answer

(x+3)(x+1)

Exercise 6.7.2

x2+6x+8

Exercise 6.7.3

x2+7x+12

Answer

(x+3)(x+4)

Exercise 6.7.4

x2+6x+5

Exercise 6.7.5

y2+8y+12

Answer

(y+6)(y+2)

Exercise 6.7.6

y25y+6

Exercise 6.7.7

y25y+4

Answer

(y4)(y1)

Exercise 6.7.8

a2+a6

Exercise 6.7.9

a2+3a4

Answer

(a+4)(a1)

Exercise 6.7.10

x2+4x21

Exercise 6.7.11

x24x21

Answer

(x7)(x+3)

Exercise 6.7.12

x2+7x+12

Exercise 6.7.13

y2+10y+16

Answer

(y+8)(y+2)

Exercise 6.7.14

x2+6x16

Exercise 6.7.15

y28y+7

Answer

(y7)(y1)

Exercise 6.7.16

y25y24

Exercise 6.7.17

a2+a30

Answer

(a+6)(a5)

Exercise 6.7.18

a23a+2

Exercise 6.7.19

a212a+20

Answer

(a10)(a2)

Exercise 6.7.20

y24y32

Exercise 6.7.21

x2+13x+42

Answer

(x+6)(x+7)

Exercise 6.7.22

x2+2x35

Exercise 6.7.23

x2+13x+40

Answer

(x+5)(x+8)

Exercise 6.7.24

y2+6y27

Exercise 6.7.25

b2+15b+56

Answer

(b+8)(b+7)

Exercise 6.7.26

3a2+24a+36

(Hint: Always search for a common factor.)

Exercise 6.7.27

4x2+12x+8

Answer

4(x+2)(x+1)

Exercise 6.7.28

2a218a+40

Exercise 6.7.29

5y270y+440

Answer

5(y214y+88)

Exercise 6.7.30

6x254x+48

Exercise 6.7.31

x3+6x2+8x

Answer

x(x+4)(x+2)

Exercise 6.7.32

x38x2+15x

Exercise 6.7.33

x4+9x3+14x2

Answer

x2(x+7)(x+2)

Exercise 6.7.34

2a3+12a2+10a

Exercise 6.7.35

4a340a2+84a

Answer

4a(a7)(a3)

Exercise 6.7.36

3xm2+33xm+54x

Exercise 6.7.37

2y2n210y2n48y2

Answer

2y2(n8)(n+3)

Exercise 6.7.38

4x442x3+144x2

Exercise 6.7.39

y5+13y4+42y3

Answer

y3(y+6)(y+7)

Exercise 6.7.40

4x2a648x2a5+252x2a4

Exercises for Review

Exercise 6.7.41

Factor 6xy+2ax3aya2.

Answer

(2xa)(3y+a)

Exercise 6.7.42

Factor 8a250.

Exercise 6.7.43

Factor 4x2+17x15.

Answer

(4x3)(x+5)


This page titled 6.7: Factoring Trinomials with Leading Coefficient 1 is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Denny Burzynski & Wade Ellis, Jr. (OpenStax CNX) .

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