7.2: Factoring trinomials of the form x² + bx + c
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Factoring with three terms, or trinomials, is the most important technique, especially in further algebra. Since factoring is a product of factors, we first look at multiplying to develop the process of factoring trinomials.
Factoring Trinomials of the Form x2+bx+c
If we multiply (x+p)(x+q), we would get
x2+px+qx+pqx2+(p+q)x+pq
Notice the two factors of the last coefficient must add up to be the middle coefficient, i.e.,
p⋅q=c and p+q=b
Hence, if we can find two numbers whose sum is b and that multiply to c, then we can split the middle term and factor by grouping.
Step 1. Find two numbers, p and q, whose sum is b and product is c.
Step 2. Rewrite the expression so that the middle term is split into two terms, p and q.
Step 3. Factor by grouping.
Step 4. Verify the factored form by finding the product.
Factor: x2+9x+18
Solution
First we identify b=9 and c=18. We ask ourselves,“ What two numbers multiply to 18 that add up to 9?”
Step 1. Find two numbers whose sum is 9 and product is 18:
p and q | Product | Sum |
---|---|---|
2,9 | 18 | 11 |
3,6 | 18 | 9 |
1,18 | 18 | 19 |
We can see from the table that 3 and 6 are the two numbers whose product is 18 and sum is 9. We use these two numbers in Step 2.
Step 2. Rewrite the expression so that the middle term is split into two terms, 3x and 6x: x2+9x+18x2+3x+6x⏟sum is 9x+18
Step 3. Factor by grouping. x2+6x+3x+18Group the first two terms and the last two terms(x2+3x)+(6x+18)Factor x from the first group and 6from the second groupx(x+3)+6(x+3)Factor the GCF (x+3)(x+3)(x+6)Factored form
Step 4. Verify the factored form by finding the product: (x+3)(x+6)FOILx2+6x+3x+18Combine like termsx2+9x+18✓ Original expression
Thus, the factored form is (x+3)(x+6).
Factor: x2−4x+3
Solution
First we identify b=−4 and c=3. We ask ourselves,“ What two numbers multiply to 3 that add up to −4?”
Step 1. Find two numbers whose sum is −4 and product is 3:
p and q | Product | Sum |
---|---|---|
1,3 | 3 | 4 |
−1,−3 | 3 | −4 |
We can see from the table that −1 and −3 are the two numbers whose product is 3 and sum is −4. We use these two numbers in Step 2.
Step 2. Rewrite the expression so that the middle term is split into two terms, −1x and −3x: x2−4x+3x2−1x−3x⏟sum is −4x+3
Step 3. Factor by grouping. x2−1x−3x+3Group the first two terms and the last two terms(x2−1x)+(−3x+3)Factor x from the first group and −3from the second groupx(x−1)−3(x−1)Factor the GCF (x−1)(x−1)(x−3)Factored form
Step 4. Verify the factored form by finding the product: (x−1)(x−3)FOILx2−1x−3x+3Combine like termsx2−4x+3✓ Original expression
Thus, the factored form is (x−1)(x+3).
Factor: x2−8x−20
Solution
First we identify b=−8 and c=−20. We ask ourselves,“ What two numbers multiply to −20 that add up to −8?”
Step 1. Find two numbers whose sum is −8 and product is −20:
p and q | Product | Sum |
---|---|---|
−4,5 | −20 | 1 |
4,−5 | −10 | −1 |
−2,10 | −20 | 8 |
2,−10 | −20 | −8 |
1,−20 | −20 | −19 |
−1,20 | −20 | 19 |
We can see from the table that 2 and −10 are the two numbers whose product is −20 and sum is −8. We use these two numbers in Step 2.
Step 2. Rewrite the expression so that the middle term is split into two terms, 2x and −10x: x2−8x−20x2+2x−10x⏟sum is −8x−20
Step 3. Factor by grouping. x2+2x−10x−20Group the first two terms and the last two terms(x2+2x)+(−10x−20)Factor x from the first group and −10from the second groupx(x+2)−10(x+2)Factor the GCF (x+2)(x+2)(x−10)Factored form
Step 4. Verify the factored form by finding the product: (x+2)(x−10)FOILx2+2x−10x−20Combine like termsx2−8x−20✓ Original expression
Thus, the factored form is (x+2)(x+10).
Factor: a2−9ab+14b2
Solution
First we identify b=−9 and c=14. We ask ourselves,“ What two numbers multiply to 14 that add up to −9?”
Step 1. Find two numbers whose sum is −9 and product is 14:
p and q | Product | Sum |
---|---|---|
2,7 | 14 | 9 |
−2,−7 | 14 | −9 |
−1,−14 | 14 | −15 |
1,14 | 14 | 15 |
We can see from the table that −2 and −7 are the two numbers whose product is 14 and sum is −9. We use these two numbers in Step 2.
Step 2. Rewrite the expression so that the middle term is split into two terms, −2ab and −7ab: a2−9ab+14b2a2−2ab−7ab⏟sum is −9ab+14b2
Step 3. Factor by grouping. a2−2ab−7ab+14b2Group the first two terms and the last two terms(a2−2ab)+(−7ab+14b2)Factor a from the first group and −7bfrom the second groupa(a−2b)−7b(a−2b)Factor the GCF (a−2b)(a−2b)(a−7b)Factored form
Step 4. Verify the factored form by finding the product: (a−2b)(a−7b)FOILa2−2ab−7ab+14b2Combine like termsa2−9ab+14b2✓ Original expression
Thus, the factored form is (a−2b)(a−7b).
There is a shortcut for factoring expressions of the type x2+bx+c. Once we identify the two numbers, p and q, whose product is c and sum is b, we can see these two numbers are the numbers in the factored form, i.e., (x+p)(x+q). We can use this shortcut only when the coefficient of x2 is 1. (We discuss when the coefficient is a number other than 1 in the next section.)
Factor: x2−7x−18
Solution
First we identify b=−7 and c=−18. We ask ourselves,“ What two numbers multiply to −18 that add up to −7?”
p and q | Product | Sum |
---|---|---|
−2,9 | −18 | 7 |
2,−9 | −18 | −7 |
−1,18 | −18 | 17 |
1,−18 | −18 | −17 |
We can see from the table that 2 and −9 are the two numbers whose product is −18 and sum is −7. We use these two numbers to rewrite the expression in factored form:
x2−7x−18(x+2)(x−9)
We can always verify the factored form by multiplying and obtaining the original expression.
Factor: m2−mn−30n2
Solution
First we identify b=−1 and c=−30. We ask ourselves,“ What two numbers multiply to −30 that add up to −1?”
p and q | Product | Sum |
---|---|---|
−2,15 | −30 | 13 |
2,−15 | −30 | −13 |
5,−6 | −30 | −1 |
−5,6 | −30 | 1 |
1,−30 | −30 | −29 |
−1,30 | −30 | 29 |
We can see from the table that 5 and −6 are the two numbers whose product is −30 and sum is −1. We use these two numbers to rewrite the expression in factored form:
m2−mn−30n2(m+5n)(m−6n)
We can always verify the factored form by multiplying and obtaining the original expression.
Factor: x2+2x+6
Solution
First we identify b=2 and c=6. We ask ourselves,“ What two numbers multiply to 6 that add up to 2?”
p and q | Product | Sum |
---|---|---|
2,3 | 6 | 5 |
−2,−3 | 6 | −5 |
1,6 | 6 | 7 |
−1,−6 | 6 | −7 |
We can see from the table that there aren’t any factors of 6 whose sum is 2. We only obtain sums with 5 and 7’s. In this case, we call this trinomial not factorable, or better yet, the trinomial is prime.
If a trinomial (or polynomial) is not factorable, then we say we the trinomial is prime.
Factoring Trinomials of the Form x2+bx+c with a Greatest Common Factor
Factoring the GCF is always the first step in factoring expressions. If all terms have a common factor, we, first, factor the GCF and then factor as usual.
Factor: 3x2−24x+45
Solution
Notice all three terms have a common factor of 3. We factor a 3 first, then factor as usual.
3x2−24x+45Factor the GCF3(x2−8x+15)
Next, we only concentrate on the expression in the parenthesis. What two numbers multiply to 15 that add up to −8?
p and q | Product | Sum |
---|---|---|
3,5 | 15 | 8 |
−3,−5 | 15 | −8 |
We can see from the table that −3 and −5 are the two numbers whose product is 15 and sum is −8. We use these two numbers to rewrite the expression in factored form:
3x2−24x+43(x−3)(x−5)
We can always verify the factored form by multiplying and obtaining the original expression.
Students tend to forget to write the GCF in the final answer. Be sure to always include the GCF in the final factored form.
Also, to factor completely, it is required the GCF is factored out of the expression. If not, then the expression is not factored completely.
The first person to use letters for unknown values was Francois Vieta in 1591 in France. He used vowels to represent variables for solving, just as codes used letters to represent an unknown message.
Factoring Trinomials of the Form x2+bx+c Homework
Factor completely.
p2+17p+72
n2−9n+8
x2−9x−10
b2+12b+32
x2+3x−70
n2−8n+15
p2+15p+54
n2−15n+56
u2−8uv+15v2
m2+2mn−8n2
x2−11xy+18y2
x2+xy−12y2
x2+4xy−12y2
5a2+60a+100
6a2+24a−192
6x2+18xy+12y2
6x2+96xy+378y2
x2+x−72
x2+x−30
x2+13x+40
b2−17b+70
x2+3x−18
a2−6a−27
p2+7p−30
m2−15mn+50n2
m2−3mn−40n2
x2+10xy+16y2
u2−9uv+14v2
x2+14xy+45y2
4x2+52x+168
5n2−45n+40
5v2+20v−25
5m2+30mn−90n2
6m2−36mn−162n2