8.3 - Factoring Quadratic Trinomials

8.3 - Factoring quadratic trinomials

Definitions: A quadratic trinomial is an expression of the form:

a x² + b x + c,

where x is a variable and a, b and c are non-zero constants. The constant a is called the leading coefficient, b is called the linear coefficient, and c is called the additive constant.

The discriminant, D, of a quadratic trinomial is defined as the quantity:

D = b² − 4 a c.

The discriminant is used to classify or discriminate among various cases of quadratic trinomials.

A perfect square number is a number that is the square of some integer. We will be interested in knowing whether or not the discriminant is a perfect square.

Example: The expression:

x² + 7 x + 6.

is a quadratic trinomial with coefficients a = 1, b = 7, and c = 6. The discriminant is D = 7² − 4 · 1 · 6 or D = 25. D is a perfect square because it is the square of 5.

Example: The expression:

2 x² + 3 x − 4.

is a quadratic trinomial with coefficients a = 2, b = 3, and c = −4. The discriminant is D = 3² − 4 · 2 · (−4) or D = 41. Notice that D is not a perfect square.

Note: If b or c is zero then the quadratic is said to be an incomplete quadratic and if a is zero then the expression isn’t even quadratic; it is linear. If any of the coefficients a, b or c is zero then the expression is no longer a trinomial and should be factored (assuming that it is factorable) by one of the simpler methods that we have studied previously:

If c = 0, then a x² + b x has a common factor, namely x.
If b = 0, then, depending on whether c is positive or negative, a x² + c is either a sum or difference of squares.
If a = 0, then b x + c is linear, not quadratic, and can only have a number as a common factor.

Assumptions for this section: We will assume that the coefficients a, b and c are all non-zero integers and that the discriminant, D, is a perfect square. If these conditions don't hold then the quadratic trinomial cannot be factored by the methods of this section but may still be factorable using the more complicated method of completing the square.

If these conditions do hold then there are three possible cases for the quadratic trinomial, each of which is factored by a different method:

discriminant D = 0 (trinomial is a perfect square)
discriminant D ≠ 0 and leading coefficient a = 1
discriminant D ≠ 0 and leading coefficient a ≠ 1 (use the grouping method)

We now study these methods.

Factoring perfect square quadratic trinomials

In this section we learn how to factor perfect square quadratic trinomials. These are quadratic trinomials that can be written as the square of some expression. Here are some examples of perfect square trinomials in unfactored and factored form:

Note that the second example can be considered to be a quadratic trinomial because we can let x play the role of the variable and let y be a constant. The discriminant, D, is always zero for perfect square trinomials. If you calculate the discriminant for the first example you find that:

and for the second example:

It is convenient to write a perfect square quadratic trinomial in the form:

a² + 2 a b + b²

Then it can be factored like this:

a² + 2 a b + b² = (a + b)²

You can verify this by distributing the right hand side of the above equation and obtaining the left hand side:

Notice that the first and last terms of the trinomial are obtained by squaring the first and last terms of the factored form:

and that the middle term of the trinomial is obtained by taking twice the product of the first and last terms of the factored form:

This leads to the following procedure for factoring a perfect square trinomial:

Verify that the trinomial is in fact a perfect square by checking that the discriminant D = 0.
Write the trinomial in the form a² + 2 a b + b².
Take the square root of the first and last terms and tentatively factor it as a² + 2 a b + b² = (a ? b)² where the ? mark means either a + sign or a − sign.
If the middle term has the same sign as the first and last terms then put a + sign like this:
a² + 2 a b + b² = (a + b)²
and if the middle term has the opposite sign of the first and last terms then put a − sign like this:
a² − 2 a b + b² = (a − b)²

Example: Factor

Verify that this is a perfect square trinomial by checking that D = 0:
Tentatively factor the trinomial like this (? means either + or −):
The middle term has the opposite sign of the first and last terms so the result is:

Example: Factor

Verify that this is a perfect square trinomial by checking that D = 0:
Tentatively factor the trinomial like this (? means either + or −):
The middle term has the opposite sign of the first and last terms so the result is:

Algebra Coach Exercises

Factoring quadratic trinomials when a = 1

In this section we learn how to factor a quadratic trinomial whose leading coefficient is a = 1:

x² + b x + c.

(We also assume that b and c are integers and that the discriminant is a perfect square.) First notice that the factors must be binomials, both of whose x terms must have coefficients of 1:

(x + •) (x + •)

Now we need to determine the two • quantities. Let's call them m and n and set the unfactored form equal to the factored form:

If we distribute the right side and then combine like terms we get:

The only way that the left side can be identical to the right side is if:

b = m + n, and c = m n.

(In words, the sum of m and n must equal b and the product of m and n must equal c.) It is easy to show that only one combination of m and n will satisfy both conditions. Also, we will see in the section on completing the square that m and n are guaranteed to be integers if the discriminant D is a perfect square.

This leads to the following procedure for factoring a quadratic trinomial a x² + b x + c when a = 1:

Verify that all of the following conditions are met for this method to work: a = 1, b and c are integers, and the discriminant D is a perfect square.
Tentatively write the trinomial in the factored form:
x² + b x + c = (x + m) (x + n)
Now find m and n. They are integers that satisfy the conditions:
m n = c, and m + n = b.
They can be found by simple trial and error: write down a list of all the pairs of integers whose product is c and then pick the one pair whose sum is b. (Notice that if c is negative then one of m and n must be positive and the other negative. If c is positive then m and n must both be positive or both negative.)

Example: Factor x² + 8 x + 12

For this trinomial a = 1, b = 8 and c = 12, and the discriminant is D = b² − 4 a c = 8² − 4 · 1 · 12 = 16, which is a perfect square, namely 4².
Tentatively write the trinomial in the factored form:
x² + 8 x + 12 = (x + m) (x + n)

Now find m and n. They are integers that satisfy the conditions:

m n = 12, and m + n = 8.

Here is a list of all the pairs of integers whose product is 12. Pick the pair whose sum is 8.

m n

12 1

6 2 ← this pair has a sum of 8

4 3

3 4 ← already checked

2 6 ← already checked

1 12 ← already checked

Thus m = 6 and n = 2 and the expression factors as:

x² + 8 x + 12 = (x + 6) (x + 2)

Note: The last 3 rows of the table duplicate the first 3 rows and don't have to be considered.

Example: Factor x² + x − 12

For this trinomial a = 1, b = 1 and c = −12, and the discriminant is D = b² − 4 a c = 1² − 4 · 1 · (−12) = 49, which is a perfect square, namely 7².
Tentatively write the trinomial in the factored form:
x² + x − 12 = (x + m) (x + n)

Now find m and n. They are integers that satisfy the conditions:

m n = −12, and m + n = 1.

Here is a list of some of the pairs of integers whose product is −12. Pick the pair whose sum is 1.

m n

12 −1

6 −2

4 −3 ← this pair has a sum of 1

… … ← need check no further

Thus m = 4 and n = −3 and the expression factors as:

x² + x − 12 = (x + 4) (x − 3)

Algebra Coach Exercises

Factoring quadratic trinomials when a ≠ 1 using the grouping method

In this section we learn how to factor a quadratic trinomial whose coefficients a, b and c can be any integers:

a x² + b x + c.

(Thus a no longer has to be equal to 1 as in the previous section but we still assume that the discriminant is a perfect square.) The method that we develop is a variation of the grouping method. Start by assuming that the quadratic trinomial can be factored like this:

a x² + b x + c = (p x + q) (r x + s).

We now set out to find p, q, r and s. Distribute on the right hand side:

a x² + b x + c = p r x² + p s x + q r x + q s.

Look at the blue and red numbers on the right hand side. Think of p r as a single number (that has factors p and r), think of p s as another number that can be factored, and so on. Then, just as in the previous case, one condition for the left side to be equal to the right side is that:

b = p s + q r.

In words, the sum of the number p s and the number q r must equal b. Unlike the previous case, the second condition is on the product a c. It is that:

a c = p r · q s = (p s) (q r).

In words, the product of the number p s and the number q r must equal the product of a and c. These two conditions together give us enough information to find the two numbers p s and q r. We will see in the section on completing the square that these two numbers are guaranteed to be integers if the discriminant D is a perfect square. Also it can be shown that they are unique.

Now use the two numbers p s and q r as a guide to split up the x term of the trinomial into two terms, like this:

a x² + b x + c = a x² + p s x + q r x + c,

or, showing the factors of a and c, like this:

a x² + b x + c = p r x² + p s x + q r x + q s.

Now make two groups on the right hand side (hence the name grouping method). The first group consists of the first two terms (shown in red) and the second group consists of the last two terms (shown in blue):

a x² + b x + c = p r x² + p s x + q r x + q s.

Factor one common factor out of the first group and a different common factor out of the last group like this:

a x² + b x + c = p x(r x + s) + q(r x + s).

This causes a new common factor of r x + s to appear, which we can also factor out, and we are done:

a x² + b x + c = (p x + q) (r x + s).

This leads to the following procedure for factoring a quadratic trinomial a x² + b x + c when a ≠ 1:

Verify that a, b and c are all integers and that the discriminant D is a perfect square.
Find two integers m and n that satisfy the conditions:
m n = a c, and m + n = b.
They can be found by simple trial and error: write down a list of all the pairs of integers whose product is a c and then pick the one pair whose sum is b. (Notice that if a c is negative then one of m and n must be positive and the other negative. If a c is positive then m and n must both be positive or both negative.)
Use m and n to split up the x term of the trinomial like this:
a x² + b x + c = a x² + m x + n x + c.
Make two groups on the right hand side, the first group consisting of the first two terms and the last group consisting of the last two terms. Factor one common factor out of the first group and another common factor out of the last group. This causes a new common factor to appear. Factor that common factor out as well and the result is the factored trinomial.

Example: Factor 6 x² + 7 x − 5

For this trinomial a = 6, b = 7 and c = −5, and the discriminant is D = b² − 4 a c = 7² − 4 · 6 · (−5) = 169, which is a perfect square, namely 13², so the conditions are met for the grouping method to work.

Find integers m and n that satisfy the conditions:

m n = a c = −30, and m + n = b = 7.

Here is a list of all the pairs of integers whose product is −30. Pick the pair whose sum is 7.

m n

30 −1

15 −2

10 −3 ← this pair has a sum of 7

6 −5

… … ← need check no further; rest of list duplicates previous rows

Use the two numbers m = 10 and n = −3 as a guide to split up the x term of the trinomial into two terms:
6 x² + 7 x − 5 = 6 x² + 10 x − 3 x − 5.
Think of this as two groups, each consisting of two terms:
6 x² + 7 x − 5 = 6 x² + 10 x − 3 x − 5.
Factor the common factor 2 x out of the first group and a − sign out of the last group:
6 x² + 7 x − 5 = 2 x(3 x + 5) − (3 x + 5).
Now factor out the new common factor 3 x + 5, and we are done:
6 x² + 7 x − 5 = (2 x − 1) (3 x + 5).

Example: Factor 8 x² + 2 x − 15

For this trinomial a = 8, b = 2 and c = −15, and the discriminant is D = b² − 4 a c = 2² − 4 · 8 · (−15) = 484, which is a perfect square, namely 22², so the conditions are met for the grouping method to work.

Find integers m and n that satisfy the conditions:

m n = a c = −120, and m + n = b = 2.

Here is a list of all the pairs of integers whose product is −120. Pick the pair whose sum is 2.

m n

120 −1

60 −2

40 −3

30 −4

24 −5

20 −6

15 −8

12 −10 ← this pair has a sum of 2

… … ← need check no further; rest of list duplicates previous rows

Use the two numbers m = 12 and n = −10 as a guide to split up the x term of the trinomial into two terms:
8 x² + 2 x − 15 = 8 x² + 12 x − 10 x − 15.
Think of this as two groups, each consisting of two terms:
8 x² + 2 x − 15 = 8 x² + 12 x − 10 x − 15.
Factor the common factor 4 x out of the first group and −5 out of the last group:
8 x² + 2 x − 15 = 4 x(2 x + 3) − 5(2 x + 3).
Now factor out the new common factor 2 x + 3, and we are done:
8 x² + 2 x − 15 = (2 x + 3) (4 x − 5).

Algebra Coach Exercises

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m	n
12	1
6	2	← this pair has a sum of 8
4	3
3	4	← already checked
2	6	← already checked
1	12	← already checked

m	n
12	−1
6	−2
4	−3	← this pair has a sum of 1
…	…	← need check no further