# 4.3: Factoring Trinomials

Learning Objectives

• Factor trinomials of the form (x^{2}+bx+c).
• Factor trinomials of higher degree.
• Factor trinomials of the form (ax^{2}+bx+c).
• Factor trinomials using the AC method.

## Factoring Trinomials Whose Leading Coefficient Is One

(Factoring Trinomials of the Form (x^{2}+bx+c))

Some trinomials of the form (x^{2}+bx+c) can be factored as a product of binomials. If a trinomial of this type factors, then we have:

(egin{aligned} x ^ { 2 } + b x + c & = ( x + m ) ( x + n ) & = x ^ { 2 } + n x + m x + m n & = x ^ { 2 } + ( n + m ) x + m n end{aligned})

This gives us

(b=n+m) and (c=mn)

In short, if the leading coefficient of a factorable trinomial is (1), then the factors of the last term must add up to the coefficient of the middle term. This observation is the key to factoring trinomials using the technique known as the trial and error (or guess and check) method18.

Example (PageIndex{1}):

Factor (x^{2}+12x+20).

Solution

We begin by writing two sets of blank parentheses. If a trinomial of this form factors, then it will factor into two linear binomial factors.

(x ^ { 2 } + 12 x + 20 = ( quad ) ( quad))

Write the factors of the first term in the first space of each set of parentheses. In this case, factor (x^{2}=x⋅x).

(x ^ { 2 } + 12 x + 20 = ( x quad ) ( xquad ))

Determine the factors of the last term whose sum equals the coefficient of the middle term. To do this, list all of the factorizations of (20) and search for factors whose sum equals (12).

(egin{aligned} 20 & = 1 cdot 20 :: ightarrow:: 1 + 20 = 21 & =color{OliveGreen}{ 2 cdot 10 :: ightarrow:: 2 + 10}color{black}{ =}color{OliveGreen}{ 12} & = 4 cdot 5 :::: ightarrow:: 4 + 5 = 9 end{aligned})

Choose (20 = 2 ⋅ 10) because (2 + 10 = 12). Write in the last term of each binomial using the factors determined in the previous step.

(x ^ { 2 } + 12 x + 20 = ( x + 2 ) ( x + 10 ))

This can be visually interpreted as follows: Check by multiplying the two binomials.

(egin{aligned} ( x + 2 ) ( x + 10 ) & = x ^ { 2 } + 10 x + 2 x + 20 & = x ^ { 2 } + 12 x + 20 ::color{Cerulean}{✓} end{aligned})

[(x+2)(x+10) onumber]

Since multiplication is commutative, the order of the factors does not matter.

(egin{aligned} x ^ { 2 } + 12 x + 20 & = ( x + 2 ) ( x + 10 ) & = ( x + 10 ) ( x + 2 ) end{aligned})

If the last term of the trinomial is positive, then either both of the constant factors must be negative or both must be positive.

Example (PageIndex{2}):

Factor (x ^ { 2 } y ^ { 2 } - 7 x y + 12).

Solution

First, factor (x^{2}y^{2}=xy⋅xy).

Next, search for factors of (12) whose sum is (−7).

In this case, choose (−3) and (−4) because ((−3)(−4)=+12) and (−3+(−4)=−7).

Check

(( x y - 3 ) ( x y - 4 ))

If the last term of the trinomial is negative, then one of its factors must be negative.

Example (PageIndex{3}):

Factor: (x ^ { 2 } - 4 x y - 12 y ^ { 2 }).

Solution

Begin by factoring the first term (x ^ { 2 } = x cdot x).

The factors of (12) are listed below. In this example, we are looking for factors whose sum is (−4).

Therefore, the coefficient of the last term can be factored as (−12=2(−6)), where (2+(−6)=−4). Because the last term has a variable factor of (y^{2}), use (−12y^{2}=2y(−6y)) and factor the trinomial as follows:

Multiply to check.

(( x + 2 y ) ( x - 6 y ))

Often our first guess will not produce a correct factorization. This process may require repeated trials. For this reason, the check is very important and is not optional.

Example (PageIndex{4}):

Factor (a ^ { 2 } + 10 a - 24).

Solution

The first term of this trinomial, (a^{2}), factors as (a⋅a).

Consider the factors of (24):

(egin{aligned} 24 & = 1 cdot 24 & = color{OliveGreen}{2 cdot 12} & = 3 cdot 8 & = color{red}{4 cdot 6} end{aligned})

Suppose we choose the factors (4) and (6) because (4 + 6 = 10), the coefficient of the middle term. Then we have the following incorrect factorization:

When we multiply to check, we find the error.

(egin{aligned} ( a + 4 ) ( a + 6 ) & = a ^ { 2 } + 6 a + 4 a + 24 & = a ^ { 2 } + 10 a color{red}{+ 24}::color{red}{✗} end{aligned})

In this case, the middle term is correct but the last term is not. Since the last term in the original expression is negative, we need to choose factors that are opposite in sign. Therefore, we must try again. This time we choose the factors (−2) and (12) because (−2+12=10).

Now the check shows that this factorization is correct.

(( a - 2 ) ( a + 12 ))

If we choose the factors wisely, then we can reduce much of the guesswork in this process. However, if a guess is not correct, do not get discouraged; just try a different set of factors. Keep in mind that some polynomials are prime. For example, consider the trinomial (x^{2}+3x+20) and the factors of (20):

(egin{aligned} 20 & = 1 cdot 20 & = 2 cdot 10 & = 4 cdot 5 end{aligned})

There are no factors of (20) whose sum is (3). Therefore, the original trinomial cannot be factored as a product of two binomials with integer coefficients. The trinomial is prime.

## Factoring Trinomials of Higher Degree

We can use the trial and error technique to factor trinomials of higher degree.

Example (PageIndex{5}):

Factor (x ^ { 4 } + 6 x ^ { 2 } + 5).

Solution

Begin by factoring the first term (x ^ { 4 } = x ^ { 2 } cdot x ^ { 2 }).

(x ^ { 4 } + 6 x ^ { 2 } + 5 = left( x ^ { 2 } quad color{Cerulean}{?} ight) left( x ^ { 2 } quad color{Cerulean}{?} ight))

Since (5) is prime and the coefficient of the middle term is positive, choose (+1) and (+5) as the factors of the last term.

(egin{aligned} x ^ { 4 } + 6 x ^ { 2 } + 5 & = left( x ^ { 2 } quad color{Cerulean}{?} ight) left( x ^ { 2 } quad color{Cerulean}{?} ight) & = left( x ^ { 2 } + 1 ight) left( x ^ { 2 } + 5 ight) end{aligned})

Notice that the variable part of the middle term is (x^{2}) and the factorization checks out.

(egin{aligned} left( x ^ { 2 } + 1 ight) left( x ^ { 2 } + 5 ight) & = x ^ { 4 } + 5 x ^ { 2 } + x ^ { 2 } + 5 & = x ^ { 4 } + 6 x ^ { 2 } + 5::color{Cerulean}{✓} end{aligned})

(left( x ^ { 2 } + 1 ight) left( x ^ { 2 } + 5 ight))

Example (PageIndex{6}):

Factor: (x ^ { 2 n } + 4 x ^ { n } - 21) where (n) is a positive integer.

Solution

Begin by factoring the first term (x ^ { 2 n } = x ^ { n } cdot x ^ { n }).

Factor (- 21 = 7 ( - 3 )) because (7 + ( - 3 ) = + 4) and write

The check is left to the reader.

Exercise (PageIndex{1})

Factor (x ^ { 6 } - x ^ { 3 } - 42).

[left( x ^ { 3 } + 6 ight) left( x ^ { 3 } - 7 ight) onumber]

## Factoring Trinomials Whose Leading Coefficient Is Not One

(Factoring Trinomials of the Form (ax^{2}+bx+c))

Factoring trinomials of the form (ax^{2}+bx+c) can be challenging because the middle term is affected by the factors of both (a) and (c). In general,

(egin{aligned} color{Cerulean}{a}color{black}{ x} ^ { 2 } + color{Cerulean}{b}color{black}{ x} + color{Cerulean}{c} & = ( p x + m ) ( q x + n ) & = p q x ^ { 2 } + p n x + q m x + m n & = color{Cerulean}{p q}color{black}{ x} ^ { 2 } + color{Cerulean}{( p n + q m ) }color{black}{x} + color{Cerulean}{m n} end{aligned})

This gives us,

(a=pq) and (b=pn+qm), where (c=mn)

In short, when the leading coefficient of a trinomial is something other than (1), there will be more to consider when determining the factors using the trial and error method. The key lies in the understanding of how the middle term is obtained. Multiply ((5x+3)(2x+3)) and carefully follow the formation of the middle term. As we have seen before, the product of the first terms of each binomial is equal to the first term of the trinomial. The middle term of the trinomial is the sum of the products of the outer and inner terms of the binomials. The product of the last terms of each binomial is equal to the last term of the trinomial. Visually, we have the following: For this reason, we need to look for products of the factors of the first and last terms whose sum is equal to the coefficient of the middle term. For example, to factor (6x^{2}+29x+35), look at the factors of (6) and (35).

(egin{aligned} 6 & = 1 cdot 635 = 1 cdot 35 & = color{OliveGreen}{2 cdot 3} quad color{black}{=}color{OliveGreen}{ 5 cdot 7} end{aligned})

The combination that produces the coefficient of the middle term is (2⋅7+3⋅5=14+15=29). Make sure that the outer terms have coefficients (2) and (7), and that the inner terms have coefficients (5) and (3). Use this information to factor the trinomial.

(egin{aligned} 6 x ^ { 2 } + 29 x + 35 & = ( 2 x quad color{Cerulean}{?}color{black}{ )} ( 3 x quad color{Cerulean}{?}color{black}{ )} & = ( 2 x + 5 ) ( 3 x + 7 ) end{aligned})

We can always check by multiplying; this is left to the reader.

Example (PageIndex{7})

Factor (5 x ^ { 2 } + 16 x y + 3 y ^ { 2 }).

Solution

Since the leading coefficient and the last term are both prime, there is only one way to factor each.

(5=1⋅5) and (3=1⋅3)

Begin by writing the factors of the first term, (5x^{2}), as follows:

(5 x ^ { 2 } + 16 x y + 3 y ^ { 2 } = ( x quad color{Cerulean}{?} color{black}{)} ( 5 x quad color{Cerulean}{?}color{black}{ )})

The middle and last term are both positive; therefore, the factors of (3) are chosen as positive numbers. In this case, the only choice is in which grouping to place these factors.

((x+y)(5x+3y)) or ((x+3y)(5x+y))

Determine which grouping is correct by multiplying each expression.

(egin{aligned} ( x + y ) ( 5 x + 3 y ) & = 5 x ^ { 2 } + 3 x y + 5 x y + 3 y ^ { 2 } & = 5 x ^ { 2 } + 8 x y + 3 y ^ { 2 } x ::color{red}{✗} ( x + 3 y ) ( 5 x + y ) & = 5 x ^ { 2 } + x y + 15 x y + 3 y ^ { 2 } & = 5 x ^ { 2 } + 16 x y + 3 y ^ { 2 } ::color{Cerulean}{✓}end{aligned})

(( x + 3 y ) ( 5 x + y ))

Example (PageIndex{8})

Factor: (18 a ^ { 2 } b ^ { 2 } - a b - 4).

Solution

First, consider the factors of the coefficients of the first and last terms.

(egin{aligned} 18 & = 1 cdot 18:quad4 =color{OliveGreen}{ 1 cdot 4} & =color{OliveGreen}{ 2 cdot 9} :::::quadcolor{black}{=} 2 cdot 2 & = 3 cdot 6 end{aligned})

We are searching for products of factors whose sum equals the coefficient of the middle term, (−1). After some thought, we can see that the sum of (8) and (−9) is (−1) and the combination that gives this follows:

Factoring begins at this point with two sets of blank parentheses.

Use (2ab) and (9ab) as factors of (18a^{2}b^{2}).

Next use the factors (1) and (4) in the correct order so that the inner and outer products are (−9ab) and (8ab) respectively.

(( 2 a b - 1 ) ( 9 a b + 4 )). The complete check is left to the reader.

It is a good practice to first factor out the GCF, if there is one. Doing this produces a trinomial factor with smaller coefficients. As we have seen, trinomials with smaller coefficients require much less effort to factor. This commonly overlooked step is worth identifying early.

Example (PageIndex{9}):

Factor (12 y ^ { 3 } - 26 y ^ { 2 } - 10 y).

Solution

Begin by factoring out the GCF.

After factoring out (2y), the coefficients of the resulting trinomial are smaller and have fewer factors. We can factor the resulting trinomial using (6=2(3)) and (5=(5)(1)). Notice that these factors can produce (−13) in two ways:

Because the last term is (−5), the correct combination requires the factors (1) and (5) to be opposite signs. Here we use (2(1) = 2) and (3(−5) = −15) because the sum is (−13) and the product of ((1)(−5) = −5).

Check.

The factor (2y) is part of the factored form of the original expression; be sure to include it in the answer.

(2 y ( 2 y - 5 ) ( 3 y + 1 ))

It is a good practice to consistently work with trinomials where the leading coefficient is positive. If the leading coefficient is negative, factor it out along with any GCF. Note that sometimes the factor will be (−1).

Example (PageIndex{10})

Factor: (- 18 x ^ { 6 } - 69 x ^ { 4 } + 12 x ^ { 2 }).

Solution

In this example, the GCF is (3x^{2}). Because the leading coefficient is negative we begin by factoring out (−3x^{2}).

At this point, factor the remaining trinomial as usual, remembering to write the (−3x^{2}) as a factor in the final answer. Use (6 = 1(6)) and (−4 = 4(−1) )because (1(−1)+6(4)=23). Therefore,

(- 3 x ^ { 2 } left( x ^ { 2 } + 4 ight) left( 6 x ^ { 2 } - 1 ight)). The check is left to the reader.

Exercise (PageIndex{2})

Factor: (- 12 a ^ { 5 } b + a ^ { 3 } b ^ { 3 } + a b ^ { 5 }).

(- a b left( 3 a ^ { 2 } - b ^ { 2 } ight) left( 4 a ^ { 2 } + b ^ { 2 } ight))

## Factoring Using the AC Method

An alternate technique for factoring trinomials, called the AC method19, makes use of the grouping method for factoring four-term polynomials. If a trinomial in the form (ax^{2}+bx+c) can be factored, then the middle term, (bx), can be replaced with two terms with coefficients whose sum is (b) and product is (ac). This substitution results in an equivalent expression with four terms that can be factored by grouping.

Example (PageIndex{11}):

Factor using the AC method: (18 x ^ { 2 } - 31 x + 6).

Solution

Here (a=18, b=-31), and (c=6).

(egin{aligned} a c & = 18 ( 6 ) & = 108 end{aligned})

Factor (108), and search for factors whose sum is (−31).

In this case, the sum of the factors (−27) and (−4) equals the middle coefficient, (−31). Therefore, (−31x=−27x−4x), and we can write

Factor the equivalent expression by grouping.

(( 2 x - 3 ) ( 9 x - 2 ))

Example (PageIndex{12}):

Factor using the AC method: (4 x ^ { 2 } y ^ { 2 } - 7 x y - 15).

Solution

Here (a=4, b= -7), and (c=-15).

Factor (−60) and search for factors whose sum is (−7).

The sum of factors (5) and (−12) equals the middle coefficient, (−7). Replace (−7xy) with (5xy−12xy).

[( 4 x y + 5 ) ( x y - 3 ).]

The check is left to the reader.

If factors of (ac) cannot be found to add up to (b) then the trinomial is prime.

## Key Takeaways

• If a trinomial of the form (x^{2}+bx+c) factors into the product of two binomials, then the coefficient of the middle term is the sum of factors of the last term.
• If a trinomial of the form (ax^{2}+bx+c) factors into the product of two binomials, then the coefficient of the middle term will be the sum of certain products of factors of the first and last terms.
• If the trinomial has a greatest common factor, then it is a best practice to first factor out the GCF before attempting to factor it into a product of binomials.
• If the leading coefficient of a trinomial is negative, then it is a best practice to first factor that negative factor out before attempting to factor the trinomial.
• Factoring is one of the more important skills required in algebra. For this reason, you should practice working as many problems as it takes to become proficient.
• Exercise (PageIndex{3})

Factor.

1. (x ^ { 2 } + 5 x - 6)
2. (x ^ { 2 } + 5 x + 6)
3. (x ^ { 2 } + 4 x - 12)
4. (x ^ { 2 } + 3 x - 18)
5. (x ^ { 2 } - 14 x + 48)
6. (x ^ { 2 } - 15 x + 54)
7. (x ^ { 2 } + 11 x - 30)
8. (x ^ { 2 } - 2 x + 24)
9. (x ^ { 2 } - 18 x + 81)
10. (x ^ { 2 } - 22 x + 121)
11. (x ^ { 2 } - x y - 20 y ^ { 2 })
12. (x ^ { 2 } + 10 x y + 9 y ^ { 2 })
13. (x ^ { 2 } y ^ { 2 } + 5 x y - 50)
14. (x ^ { 2 } y ^ { 2 } - 16 x y + 48)
15. (a ^ { 2 } - 6 a b - 72 b ^ { 2 })
16. (a ^ { 2 } - 21 a b + 80 b ^ { 2 })
17. (u ^ { 2 } + 14 u v - 32 v ^ { 2 })
18. (m ^ { 2 } + 7 m n - 98 n ^ { 2 })
19. (( x + y ) ^ { 2 } - 2 ( x + y ) - 8)
20. (( x - y ) ^ { 2 } - 2 ( x - y ) - 15)
21. (x ^ { 4 } - 7 x ^ { 2 } - 8)
22. (x ^ { 4 } + 13 x ^ { 2 } + 30)
23. (x ^ { 4 } - 8 x ^ { 2 } - 48)
24. (x ^ { 4 } + 25 x ^ { 2 } + 24)
25. (y ^ { 4 } - 20 y ^ { 2 } + 100)
26. (y ^ { 4 } + 14 y ^ { 2 } + 49)
27. (x ^ { 4 } + 3 x ^ { 2 } y ^ { 2 } + 2 y ^ { 4 })
28. (x ^ { 4 } - 8 x ^ { 2 } y ^ { 2 } + 15 y ^ { 4 })
29. (a ^ { 4 } b ^ { 4 } - 4 a ^ { 2 } b ^ { 2 } + 4)
30. (a ^ { 4 } + 6 a ^ { 2 } b ^ { 2 } + 9 b ^ { 4 })
31. (x ^ { 6 } - 18 x ^ { 3 } - 40)
32. (x ^ { 6 } + 18 x ^ { 3 } + 45)
33. (x ^ { 6 } - x ^ { 3 } y ^ { 3 } - 6 y ^ { 6 })
34. (x ^ { 6 } + x ^ { 3 } y ^ { 3 } - 20 y ^ { 6 })
35. (x ^ { 6 } y ^ { 6 } + 2 x ^ { 3 } y ^ { 3 } - 15)
36. (x ^ { 6 } y ^ { 6 } + 16 x ^ { 3 } y ^ { 3 } + 48)
37. (x ^ { 2 n } + 12 x ^ { n } + 32)
38. (x ^ { 2 n } + 41 x ^ { n } + 40)
39. (x ^ { 2 n } + 2 a x ^ { n } + a ^ { 2 })
40. (x ^ { 2 n } - 2 a x ^ { n } + a ^ { 2 })

1. (( x - 1 ) ( x + 6 ))

3. (( x - 2 ) ( x + 6 ))

5. (( x - 6 ) ( x - 8 ))

7. Prime

9. (( x - 9 ) ^ { 2 })

11. (( x - 5 y ) ( x + 4 y ))

13. (( x y - 5 ) ( x y + 10 ))

15. (( a + 6 b ) ( a - 12 b ))

17. (( u - 2 v ) ( u + 16 v ))

19. (( x + y - 4 ) ( x + y + 2 ))

21. (left( x ^ { 2 } - 8 ight) left( x ^ { 2 } + 1 ight))

23. (left( x ^ { 2 } + 4 ight) left( x ^ { 2 } - 12 ight))

25. (left( y ^ { 2 } - 10 ight) ^ { 2 })

27. (left( x ^ { 2 } + y ^ { 2 } ight) left( x ^ { 2 } + 2 y ^ { 2 } ight))

29. (left( a ^ { 2 } b ^ { 2 } - 2 ight) ^ { 2 })

31. (left( x ^ { 3 } - 20 ight) left( x ^ { 3 } + 2 ight))

33. (left( x ^ { 3 } + 2 y ^ { 3 } ight) left( x ^ { 3 } - 3 y ^ { 3 } ight))

35. (left( x ^ { 3 } y ^ { 3 } - 3 ight) left( x ^ { 3 } y ^ { 3 } + 5 ight))

37. (left( x ^ { n } + 4 ight) left( x ^ { n } + 8 ight))

39. (left( x ^ { n } + a ight) ^ { 2 })

Exercise (PageIndex{4})

Factor.

1. (3 x ^ { 2 } + 20 x - 7)
2. (2 x ^ { 2 } - 9 x - 5)
3. (6 a ^ { 2 } + 13 a + 6)
4. (4 a ^ { 2 } + 11 a + 6)
5. (6 x ^ { 2 } + 7 x - 10)
6. (4 x ^ { 2 } - 25 x + 6)
7. (24 y ^ { 2 } - 35 y + 4)
8. (10 y ^ { 2 } - 23 y + 12)
9. (14 x ^ { 2 } - 11 x + 9)
10. (9 x ^ { 2 } + 6 x + 8)
11. (4 x ^ { 2 } - 28 x + 49)
12. (36 x ^ { 2 } - 60 x + 25)
13. (27 x ^ { 2 } - 6 x - 8)
14. (24 x ^ { 2 } + 17 x - 20)
15. (6 x ^ { 2 } + 23 x y - 4 y ^ { 2 })
16. (10 x ^ { 2 } - 21 x y - 27 y ^ { 2 })
17. (8 a ^ { 2 } b ^ { 2 } - 18 a b + 9)
18. (12 a ^ { 2 } b ^ { 2 } - a b - 20)
19. (8 u ^ { 2 } - 26 u v + 15 v ^ { 2 })
20. (24 m ^ { 2 } - 26 m n + 5 n ^ { 2 })
21. (4 a ^ { 2 } - 12 a b + 9 b ^ { 2 })
22. (16 a ^ { 2 } + 40 a b + 25 b ^ { 2 })
23. (5 ( x + y ) ^ { 2 } - 9 ( x + y ) + 4)
24. (7 ( x - y ) ^ { 2 } + 15 ( x - y ) - 18)
25. (7 x ^ { 4 } - 22 x ^ { 2 } + 3)
26. (5 x ^ { 4 } - 41 x ^ { 2 } + 8)
27. (4 y ^ { 6 } - 3 y ^ { 3 } - 10)
28. (12 y ^ { 6 } + 4 y ^ { 3 } - 5)
29. (5 a ^ { 4 } b ^ { 4 } - a ^ { 2 } b ^ { 2 } - 18)
30. (21 a ^ { 4 } b ^ { 4 } + 5 a ^ { 2 } b ^ { 2 } - 4)
31. (6 x ^ { 6 } y ^ { 6 } + 17 x ^ { 3 } y ^ { 3 } + 10)
32. (16 x ^ { 6 } y ^ { 6 } + 46 x ^ { 3 } y ^ { 3 } + 15)
33. (8 x ^ { 2 n } - 10 x ^ { n } - 25)
34. (30 x ^ { 2 n } - 11 x ^ { n } - 6)
35. (36 x ^ { 2 n } + 12 a x ^ { n } + a ^ { 2 })
36. (9 x ^ { 2 n } - 12 a x ^ { n } + 4 a ^ { 2 })
37. (- 3 x ^ { 2 } + 14 x + 5)
38. (- 2 x ^ { 2 } + 13 x - 20)
39. (- x ^ { 2 } - 10 x + 24)
40. (- x ^ { 2 } + 8 x + 48)
41. (54 - 12 x - 2 x ^ { 2 })
42. (60 + 5 x - 5 x ^ { 2 })
43. (4 x ^ { 3 } + 16 x ^ { 2 } + 20 x)
44. (2 x ^ { 4 } - 12 x ^ { 3 } + 14 x ^ { 2 })
45. (2 x ^ { 3 } - 8 x ^ { 2 } y - 24 x y ^ { 2 })
46. (6 x ^ { 3 } - 9 x ^ { 2 } y - 6 x y ^ { 2 })
47. (4 a ^ { 3 } b - 4 a ^ { 2 } b ^ { 2 } - 24 a b ^ { 3 })
48. (15 a ^ { 4 } b - 33 a ^ { 3 } b ^ { 2 } + 6 a ^ { 2 } b ^ { 3 })
49. (3 x ^ { 5 } y + 30 x ^ { 3 } y ^ { 3 } + 75 x y ^ { 5 })
50. (45 x ^ { 5 } y ^ { 2 } - 60 x ^ { 3 } y ^ { 4 } + 20 x y ^ { 6 })

1. (( 3 x - 1 ) ( x + 7 ))

3. (( 2 a + 3 ) ( 3 a + 2 ))

5. (( 6 x - 5 ) ( x + 2 ))

7. (( 8 y - 1 ) ( 3 y - 4 ))

9. Prime

11. (( 2 x - 7 ) ^ { 2 })

13. (( 9 x + 4 ) ( 3 x - 2 ))

15. (( 6 x - y ) ( x + 4 y ))

17. (( 4 a b - 3 ) ( 2 a b - 3 ))

19. (( 2 u - 5 v ) ( 4 u - 3 v ))

21. (( 2 a - 3 b ) ^ { 2 })

23. (( x + y - 1 ) ( 5 x + 5 y - 4 ))

25. (left( x ^ { 2 } - 3 ight) left( 7 x ^ { 2 } - 1 ight))

27. (left( y ^ { 3 } - 2 ight) left( 4 y ^ { 3 } + 5 ight))

29. (left( a ^ { 2 } b ^ { 2 } - 2 ight) left( 5 a ^ { 2 } b ^ { 2 } + 9 ight))

31. (left( 6 x ^ { 3 } y ^ { 3 } + 5 ight) left( x ^ { 3 } y ^ { 3 } + 2 ight))

33. (left( 2 x ^ { n } - 5 ight) left( 4 x ^ { n } + 5 ight))

35. (left( 6 x ^ { n } + a ight) ^ { 2 })

37. (- ( x - 5 ) ( 3 x + 1 ))

39. (- ( x - 2 ) ( x + 12 ))

41. (- 2 ( x - 3 ) ( x + 9 ))

43. (4 x left( x ^ { 2 } + 4 x + 5 ight))

45. (2 x ( x + 2 y ) ( x - 6 y ))

47. (4 a b ( a - 3 b ) ( a + 2 b ))

49. (3 x y left( x ^ { 2 } + 5 y ^ { 2 } ight) ^ { 2 })

Exercise (PageIndex{5})

Factor.

1. (4 - 25 x ^ { 2 })
2. (8 x ^ { 3 } - y ^ { 3 })
3. (9 x ^ { 2 } - 12 x y + 4 y ^ { 2 })
4. (30 a ^ { 2 } - 57 a b - 6 b ^ { 2 })
5. (10 a ^ { 2 } - 5 a - 6 a b + 3 b)
6. (3 x ^ { 3 } - 4 x ^ { 2 } + 9 x - 12)
7. (x ^ { 2 } + 4 y ^ { 2 })
8. (x ^ { 2 } - x + 2)
9. (15 a ^ { 3 } b ^ { 2 } + 6 a ^ { 2 } b ^ { 3 } - 3 a b ^ { 4 })
10. (54 x ^ { 2 } - 63 x)

1. (( 2 - 5 x ) ( 2 + 5 x ))

3. (( 3 x - 2 y ) ^ { 2 })

5. (( 2 a - 1 ) ( 5 a - 3 b ))

7. (3 a b ^ { 2 } left( 5 a ^ { 2 } + 2 a b - b ^ { 2 } ight))

Exercise (PageIndex{6})

1. Create your own trinomial of the form (ax^{2} + bx + c) that factors. Share it, along with the solution, on the discussion board.
2. Create a trinomial of the form (ax^{2} + bx + c) that does not factor and share it along with the reason why it does not factor.

## Footnotes

18Describes the method of factoring a trinomial by systematically checking factors to see if their product is the original trinomial.

19Method used for factoring trinomials by replacing the middle term with two terms that allow us to factor the resulting four-term polynomial by grouping.

## Questions

Factor each of the following trinomials.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. ## How to factor trinomials with a leading coefficient of 1

Let’s walk through the following steps to factor x 2 + 7x + 12:

• Comparing x 2 + 7x + 12 with the standard form of ax 2 + bx + c, we get, a = 1, b = 7, and c = 12
• Find the paired factors of c such that their sum is equal to b. The pair factor of 12 are (1, 12), (2, 6), and (3, 4). Therefore, the suitable pair is 3 and 4.
• In separate brackets, add each number of the pair to x to get (x + 3) and (x + 4).
• Write the two binomials side by side to get the factored result as

## PCC SLC Math Resources

Factoring a trinomial of form (x^2+bx+c ext<,>) where (b) and (c) are integers, is essentially the reversal of a FOIL process. For this reason, we can develop a strategy by investigating a FOIL expansion. Let's expand ((x+4)(x+6) ext<.>)

egin (x+4)(x+6)amp=x^2+6x+4x+24 amp=x^2+10x+24 end

Let's observe that the linear coefficient, (10 ext<,>) is the sum of (4) and (6) whereas the constant term, (24 ext<,>) is the product of (4) and (6 ext<.>) Since we expanded ((x+4)(x+6) ext<,>) we can infer that finding the constants in a factorization of trinomials of form (x^2+bx+c) is dependent upon determining two numbers that sum to (b) and multiply to (c ext<.>)

From the factor pairs and sums shown in Table 1.2.1, we can infer the following factorizations.

Just as noteworthy, since the factor pairs shown in Table 1.2.1 form an exhaustive list, we cannot factor any trinomial of form (x^2+bx+6) where b is not one of the four numbers shown in the sum column. Such trinomials are said to be prime. For example, (x^2+3x+6) is prime.

Let's factor (x^2-8x+15 ext<.>) Our first task is to determine a factor pair that multiplies to (15) and adds to (-8 ext<.>) Since the product is positive and the sum is negative, we are searching for two negative numbers. The pair that works is (-3) and (-5 ext<.>) This gives us:

Now let's factor (w^2-4w-12 ext<.>) Since our factor pair needs to multiply to a negative value ((-12) ext<,>) one the numbers in the pair must be positive and the other negative. Let's note that (6) and (2) multiply to 12 and have a difference of (4 ext<.>) This can be helpful in determining that the factor pair that multiplies to (-12) and adds to (-4) is (-6) and (2 ext<.>) This gives us

We also could have written the factorization in the reverse order:

###### 2 Factoring by grouping

The expressions that are factored in this set of examples are not trinomials - they all have one too many terms. The reason we are exploring this skill is that it is a useful process when factoring trinomials where the leading coefficients isn't (1 ext<.>)

Let's consider the expression (10xy+15x-6y^2-9y ext<.>) The factoring by grouping method begins by considering only the first two terms and factoring from them any and all common factors. In this case we could factor out (5x) leaving the residual factor ((2y+3) ext<.>) We now consider the final two terms. Our goal is to factor out something that also leaves the residual factor ((2y+3) ext<.>) While it's true that we could factor out (3y ext<,>) that would leave behind ((-2y-3) ext<.>) What we want to factor out is (-3y) so that the residual factor is the desired ((2y+3) ext<.>) Altogether, then, the first step in the process is:

egin 10xy+15x-6xy-9y=5x(2y+3)-3y(2y+3) ext <.>end

The expression now has two terms, with the terms delineated by the subtraction sign. Each of the terms has a factor of ((2y+3) ext<.>) Just like we can factor, say, (z) from the expression (5xz-3yz) resulting in (z(5x-3y) ext<,>) we can factor ((2y+3)) for the expression (5x(2y+3)-3y(2y+3)) resulting in ((2y+3)(5x-3y) ext<.>)

Let's see the process in total:

egin 10xy+15x-6y^2-9yamp=5x(2y+3)-3y(2y+3) amp=(2y+3)(5x-3y) end

We can use FOIL to check our result.

Let's factor (6x^2y-4x^2+12y-8 ext<.>) We begin by observing that (2x^2) can be factored from the first two terms while (4) can be factored from the final two terms. Heads up! One of the most common errors made during this process is to forget to put a plus sign in front of the factor of (4 ext<.>) Written correctly, our factorization process is:

egin 6x^2y-4x^2+12y-8amp=2x^2(3y-2)+4(3y-2) amp=(3y-2)(2x^2+4) end

One more example. Let's factor (7w^2-2w+7w-2 ext<.>) We can see that (w) can be factored from the first two terms, but there seems to be nothing that can be factored from the final two terms. When we factor (w) from the first two terms, the residual factor is ((7w-2)) which contains the two terms at the rear of our original expression. The resolution of our dilemma, then, is to factor (1) away from the final two terms. We need to remember to put a plus sign in from of that factor of (1 ext<.>) Going through the complete process we have:

egin 7w^2-2w+7w-2amp=w(7w-2)+1(7w-2) amp=(7w-2)(w+1) end

Before closing, let's note that (7w^2-2w+7w-2) simplifies to (7w^2+5w-2 ext<.>) This gives us some insight into how the factoring by grouping process will be useful when factoring trinomials where the leading coefficient isn't (1 ext<.>)

###### 3 Factoring trinomials where the leading coefficients are not (1)

When factoring trinomials it is important to first look for factors common to all three terms. While we won't lead with an example of this type, it's always good to remind ourselves of this. Some of the problems in the your next problem set are all but impossible to resolve if you omit this step.

That said, our first few examples are going to deal with trinomials of form (ax^2+bx+c) where (a eq1 ext<.>) We are going to use what is known as the (ac)-method to factor. The initial task is to determine two factors of the product (ac) that sum to (b ext<.>) We will then rewrite the expression using that factor pair to split the linear term into two terms and factor the result by grouping. Hopefully that will make more sense when you see some examples!

Let's start by factoring (6x^2+7x-5 ext<.>) Our initial task is to find a factor pair of (-30) ((ac)) that sums to (7 ext<.>) The pair that works is (-3) and (10 ext<.>) Let's precede with the strategy outlined above.

egin 6x^2+7x-5amp=6x^2-3x+10x-5 amp=3x(2x-1)+5(2x-1) amp=(2x-1)(3x+5) end

Let's now factor (3x^2+16xy-12y^2 ext<.>) Our first objective is to determine two factors of (-36) that sum to (16 ext<.>) The pair that works is (18) and (-2 ext<.>) Proceeding to our factoring by grouping we have:

egin 3x^2+16xy-12y^2amp=3x^2+18xy-2xy-12y^2 amp=3x(x+6y)-2y(x+6y) amp=(x+6y)(3x-2y) end

Let's try one more: (24w^4-42w^3z+9w^2z^2 ext<.>) The first thing we should notice is that there are common factors to all three terms. Three evenly divides into each term, and each term contains at least two factors of (w ext<,>) so we can factor (3w^2) for the expression.

egin 24w^4-42w^3z+9w^2z^2=3w^2(8w^2-14wz+3z^2) end

Turning our attention to the expression in the last set of parentheses, we need to determine a factor pair of 24 that sums to -14. The factor pair is (-12) and (-2 ext<.>) This gives us:

egin 24w^4-42w^3z+9w^2z^2amp=3w^2(8w^2-14wz+3z^2) amp=3w^2(8w^2-12wz-2wz+3z^2) amp=3w^2[4w(2w-3z)-z(2w-3z)] amp=3w^2(2w-3z)(4w-z) end

### Subsection 1.2.2 Practice Exercises

###### 1 Factoring trinomials with leading coefficients of (1)

Factor each trinomial. Check your answer by expanding the factorization. If the trinomial cannot be factored, state that it is prime.

1. (x^2+6x+5)
2. (t^2-6t+8)
3. (x^2+8x+4)
4. (y^2-14y-32)
5. (x^2+6x-20)
6. (x^2+15x-100)
7. (w^2-18w+45)
8. (x^2+21x+80)
9. (b^2-9b-400)
1. (x^2+6x+5=(x+2)(x+3))
2. (t^2-6t+8=(t-2)(t-4))
3. (x^2+8x+4) is prime.
4. (y^2-14y-32=(y-16)(y+2))
5. (x^2+6x-20) is prime.
6. (x^2+15x-100=(x+20)(x-5))
7. (w^2-18w+45=(w-15)(w-3))
8. (x^2+21x+80) is prime
9. (b^2-9b-400=(b-25)(b+16))
###### 2 Factoring by grouping

Factor each of the following expressions. Check each result using the FOIL technique.

## Factoring Trinomials

After you look for common factors, if you have a trinomial that looks something like this: ,then you want to the following process:

1) There are no common factors, so we set up our parenthesis. Next we factor the first term. .This is the only choice, so we know this is the right set and we can list these in the front of each parenthesis Write our set of factors -2(-1) one in the first parenthesis and
one in the second. Remember to include the signs.We can
Final answer:  . This matches the original, so we factored correctly. There are no common factors, so we set up our parenthesis. Next we factor the first term. . This is the only choice, so we know this is the right set and we can list these in the front of each parenthesis . This matches the original, so we factored correctly.  There is a common factor of 2, so we must factor that out first. Next we factor the first term. . Notice of that we just
2 bring the 2 down, but it won't affect the rest of the factoring.  , so this is our set. Put them in the parenthesis.We can check our answer.
Final answer:  This matches the original, so we factored correctly.

1) 2) 3) 4) 5) ## 6.2 Factor Trinomials

Find all the factors of 72.
If you missed this problem, review Example 1.2.

### Factor Trinomials of the Form x 2 + b x + c x 2 + b x + c

You have already learned how to multiply binomials using FOIL . Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.

 Foil to find the product. Factor the GCF from the middle terms. Our trinomial is of the form x 2 + b x + c . x 2 + b x + c .

### Example 6.9

#### Solution

Let’s summarize the steps we used to find the factors.

### How To

#### Factor trinomials of the form x 2 + b x + c . x 2 + b x + c .

In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.

How do you get a positive product and a negative sum? We use two negative numbers.

### Example 6.10

#### Solution

Now, what if the last term in the trinomial is negative? Think about FOIL . The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.

How do you get a negative product and a positive sum? We use one positive and one negative number.

When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.

### Example 6.12

#### Solution

We need r in the first term of each binomial and s in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

Factor: a 2 − 11 a b + 10 b 2 . a 2 − 11 a b + 10 b 2 .

Factor: m 2 − 13 m n + 12 n 2 . m 2 − 13 m n + 12 n 2 .

Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.

### Example 6.13

Factor: u 2 − 9 u v − 12 v 2 . u 2 − 9 u v − 12 v 2 .

#### Solution

We need u in the first term of each binomial and v in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

Note there are no factor pairs that give us −9 −9 as a sum. The trinomial is prime.

Factor: x 2 − 7 x y − 10 y 2 . x 2 − 7 x y − 10 y 2 .

Factor: p 2 + 15 p q + 20 q 2 . p 2 + 15 p q + 20 q 2 .

Let’s summarize the method we just developed to factor trinomials of the form x 2 + b x + c . x 2 + b x + c .

### Strategy for Factoring Trinomials of the Form x 2 + b x + c x 2 + b x + c

When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.

Notice that, in the case when m and n have opposite signs, the sign of the one with the larger absolute value matches the sign of b.

### Factor Trinomials of the form ax 2 + bx + c using Trial and Error

Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form a x 2 + b x + c . a x 2 + b x + c .

Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works.

### Example 6.14

Factor completely: 4 x 3 + 16 x 2 − 20 x . 4 x 3 + 16 x 2 − 20 x .

#### Solution

Factor completely: 5 x 3 + 15 x 2 − 20 x . 5 x 3 + 15 x 2 − 20 x .

Factor completely: 6 y 3 + 18 y 2 − 60 y . 6 y 3 + 18 y 2 − 60 y .

What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Let’s factor the trinomial 3 x 2 + 5 x + 2 . 3 x 2 + 5 x + 2 .

From our earlier work, we expect this will factor into two binomials.

We know the first terms of the binomial factors will multiply to give us 3 x 2 . 3 x 2 . The only factors of 3 x 2 3 x 2 are 1 x , 3 x . 1 x , 3 x . We can place them in the binomials.

Check: Does 1 x · 3 x = 3 x 2 ? 1 x · 3 x = 3 x 2 ?

We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1, 2. But we now have two cases to consider as it will make a difference if we write 1, 2 or 2, 1.

Which factors are correct? To decide that, we multiply the inner and outer terms.

Our result of the factoring is:

### Example 6.15

#### How to Factor a Trinomial Using Trial and Error

Factor completely using trial and error: 3 y 2 + 22 y + 7 . 3 y 2 + 22 y + 7 .

#### Solution

Factor completely using trial and error: 2 a 2 + 5 a + 3 . 2 a 2 + 5 a + 3 .

Factor completely using trial and error: 4 b 2 + 5 b + 1 . 4 b 2 + 5 b + 1 .

### How To

#### Factor trinomials of the form a x 2 + b x + c a x 2 + b x + c using trial and error.

1. Step 1. Write the trinomial in descending order of degrees as needed.
2. Step 2. Factor any GCF.
3. Step 3. Find all the factor pairs of the first term.
4. Step 4. Find all the factor pairs of the third term.
5. Step 5. Test all the possible combinations of the factors until the correct product is found.
6. Step 6. Check by multiplying.

Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

### Example 6.16

Factor completely using trial and error: 6 b 2 − 13 b + 5 . 6 b 2 − 13 b + 5 .

#### Solution

 The trinomial is already in descending order. Find the factors of the first term. Find the factors of the last term. Consider the signs. Since the last term, 5, is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.

Consider all the combinations of factors.

Factor completely using trial and error: 8 x 2 − 14 x + 3 . 8 x 2 − 14 x + 3 .

Factor completely using trial and error: 10 y 2 − 37 y + 7 . 10 y 2 − 37 y + 7 .

When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

### Example 6.17

Factor completely using trial and error: 18 x 2 − 37 x y + 15 y 2 . 18 x 2 − 37 x y + 15 y 2 .

#### Solution

 The trinomial is already in descending order. Find the factors of the first term. Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative factors.

Consider all the combinations of factors.

Factor completely using trial and error 18 x 2 − 3 x y − 10 y 2 . 18 x 2 − 3 x y − 10 y 2 .

Factor completely using trial and error: 30 x 2 − 53 x y − 21 y 2 . 30 x 2 − 53 x y − 21 y 2 .

Don’t forget to look for a GCF first and remember if the leading coefficient is negative, so is the GCF.

### Example 6.18

Factor completely using trial and error: −10 y 4 − 55 y 3 − 60 y 2 . −10 y 4 − 55 y 3 − 60 y 2 .

#### Solution

Consider all the combinations.

 The correct factors are those whose product is the original trinomial. Remember to include the factor − 5 y 2 . − 5 y 2 . − 5 y 2 ( y + 4 ) ( 2 y + 3 ) − 5 y 2 ( y + 4 ) ( 2 y + 3 ) Check by multiplying: − 5 y 2 ( y + 4 ) ( 2 y + 3 ) − 5 y 2 ( 2 y 2 + 8 y + 3 y + 12 ) − 10 y 4 − 55 y 3 − 60 y 2 ✓ − 5 y 2 ( y + 4 ) ( 2 y + 3 ) − 5 y 2 ( 2 y 2 + 8 y + 3 y + 12 ) − 10 y 4 − 55 y 3 − 60 y 2 ✓

Factor completely using trial and error: 15 n 3 − 85 n 2 + 100 n . 15 n 3 − 85 n 2 + 100 n .

Factor completely using trial and error: 56 q 3 + 320 q 2 − 96 q . 56 q 3 + 320 q 2 − 96 q .

### Example 6.19

#### How to Factor Trinomials using the “ac” Method

Factor using the ‘ac’ method: 6 x 2 + 7 x + 2 . 6 x 2 + 7 x + 2 .

#### Solution

Factor using the ‘ac’ method: 6 x 2 + 13 x + 2 . 6 x 2 + 13 x + 2 .

Factor using the ‘ac’ method: 4 y 2 + 8 y + 3 . 4 y 2 + 8 y + 3 .

The “ac” method is summarized here.

### How To

#### Factor trinomials of the form a x 2 + b x + c a x 2 + b x + c using the “ac” method.

1. Step 1. Factor any GCF.
2. Step 2. Find the product ac.
3. Step 3. Find two numbers m and n that:
Multiply to a c m · n = a · c Add to b m + n = b a x 2 + b x + c Multiply to a c m · n = a · c Add to b m + n = b a x 2 + b x + c
4. Step 4. Split the middle term using m and n. a x 2 + m x + n x + c a x 2 + m x + n x + c
5. Step 5. Factor by grouping.
6. Step 6. Check by multiplying the factors.

Don’t forget to look for a common factor!

### Example 6.20

Factor using the ‘ac’ method: 10 y 2 − 55 y + 70 . 10 y 2 − 55 y + 70 .

#### Solution

 Is there a greatest common factor? Yes. The GCF is 5. Factor it. The trinomial inside the parentheses has a leading coefficient that is not 1. Find the product a c . a c . a c = 28 a c = 28 Find two numbers that multiply to a c a c ( −4 ) ( −7 ) = 28 ( −4 ) ( −7 ) = 28 and add to b. −4 + ( −7 ) = −11 −4 + ( −7 ) = −11 Split the middle term. Factor the trinomial by grouping. Check by multiplying all three factors. 5 ( y − 2 ) ( 2 y − 7 ) 5 ( 2 y 2 − 7 y − 4 y + 14 ) 5 ( 2 y 2 − 11 y + 14 ) 10 y 2 − 55 y + 70 ✓ 5 ( y − 2 ) ( 2 y − 7 ) 5 ( 2 y 2 − 7 y − 4 y + 14 ) 5 ( 2 y 2 − 11 y + 14 ) 10 y 2 − 55 y + 70 ✓

Factor using the ‘ac’ method: 16 x 2 − 32 x + 12 . 16 x 2 − 32 x + 12 .

Factor using the ‘ac’ method: 18 w 2 − 39 w + 18 . 18 w 2 − 39 w + 18 .

### Example 6.21

Factor by substitution: x 4 − 4 x 2 − 5. x 4 − 4 x 2 − 5.

#### Solution

Factor by substitution: h 4 + 4 h 2 − 12 . h 4 + 4 h 2 − 12 .

Factor by substitution: y 4 − y 2 − 20 . y 4 − y 2 − 20 .

Sometimes the expression to be substituted is not a monomial.

### Example 6.22

Factor by substitution: ( x − 2 ) 2 + 7 ( x − 2 ) + 12 ( x − 2 ) 2 + 7 ( x − 2 ) + 12

#### Solution

Factor by substitution: ( x − 5 ) 2 + 6 ( x − 5 ) + 8 . ( x − 5 ) 2 + 6 ( x − 5 ) + 8 .

Factor by substitution: ( y − 4 ) 2 + 8 ( y − 4 ) + 15 . ( y − 4 ) 2 + 8 ( y − 4 ) + 15 .

### Media

Access this online resource for additional instruction and practice with factoring.

### Section 6.2 Exercises

#### Practice Makes Perfect

Factor Trinomials of the Form x 2 + b x + c x 2 + b x + c

In the following exercises, factor each trinomial of the form x 2 + b x + c . x 2 + b x + c .

In the following exercises, factor each trinomial of the form x 2 + b x y + c y 2 . x 2 + b x y + c y 2 .

Factor Trinomials of the Form a x 2 + b x + c a x 2 + b x + c Using Trial and Error

In the following exercises, factor completely using trial and error.

Factor Trinomials of the Form a x 2 + b x + c a x 2 + b x + c using the ‘ac’ Method

In the following exercises, factor using the ‘ac’ method.

Factor Using Substitution

In the following exercises, factor using substitution.

( 5 y − 1 ) 2 − 3 ( 5 y − 1 ) − 18 ( 5 y − 1 ) 2 − 3 ( 5 y − 1 ) − 18

Mixed Practice

In the following exercises, factor each expression using any method.

#### Writing Exercises

List, in order, all the steps you take when using the “ac” method to factor a trinomial of the form a x 2 + b x + c . a x 2 + b x + c .

How is the “ac” method similar to the “undo FOIL” method? How is it different?

#### Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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## 4.3: Factoring Trinomials

After studying this lesson, you will be able to:

Steps of Factoring:

2. Look at the number of terms:

• 2 Terms: Look for the Difference of 2 Squares
• 3 Terms: Factor the Trinomial
• 4 Terms: Factor by Grouping

This lesson will concentrate on the second step of factoring:Factoring Trinomials.

**When there are 3 terms, we are factoring trinomials. Don'tforget to look for a GCF first.**

Factoring trinomials often requires some trial and error.Don't get frustrated. Try all possible combinations. In theprevious problems, the first term has not had a coefficient. Wewill now look at problems that do have coefficients in the firstterm. This adds another level of trial and error or"guessing".

One thing that will make the "guessing" moreaccurate is to look for a prime number in the first term or theconstant term. Remember, a prime number only has 2 factors. 1and itself. If the coefficient of the first term or the constantterm is prime, start there and "lock in" those factors.

This is a trinomial (has 3 terms). There is no GCF other thanone. So, we start with 2 parentheses:

Using our signs rules, we can determine the signs for thefactors. The constant term is negative so we know the signs willbe different. Keep this in mind.

1 st : Since the coefficient of the constant term is prime(5), we will start with the constant term. Find the factors ofthe constant term. The factors of 5 are 1 and 5 . These go in thelast positions. We won't put the signs in yet because we aren'tsure where they go.

2 nd : Find the factors of the first term. The factors of 6x 2 are 1x, 6x and 2x, 3x. Remember, we need theinside/outside combination to add up to the middle term which is-13x. This time we don't just consider the factors of theconstant term because the first term also had factors. Here'swhere the guessing comes in. Let's try the factors 2x,3x and seewhat happens.

Notice we still didn't put in the signs. Let's checkthe inside/outside combination. If we multiply inside, 1 times 3xgives us 3x. Multiplying outside 2x times 5 gives us 10x. Nowremember, we have to have different signs. On the inside/outsidecombination we have 3x and 10x. Using different signs, we cannotmake the combination equal the middle term. We resort toguessing. Let's reverse the 2x and 3x and see what happens.

Notice we still didn't put in the signs. Let's checkthe inside/outside combination. If we multiply inside, 1 times 2xgives us 2x. Multiplying outside 3x times 5 gives us 15x. Nowremember, we have to have different signs. On the inside/outsidecombination we have 2x and 15x. If we make the 15x negative andthe 2x positive we will have the combination of -13x. which isour middle term.

Check by using FOIL (3x + 1) (2x - 5) 6x 2 - 15x + 2x - 5 which is6x 2 - 13x - 5