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Factoring Quadratic Trinomials with Nonunit Leading Coefficients

Welcome back. In the previous lesson, you established the first rule of complete factoring: extract any greatest common factor before doing anything else. That rule remains the opening move today, but now the remaining expression may be a quadratic trinomial whose leading coefficient is not $1$ , such as $6 x^{2} - x - 15$ .

The goal is to factor trinomials of the form

a x^{2} + b x + c

where $a$ is an integer other than $1$ , into factors with integer coefficients whenever possible. The reliable method is usually called the $a c$ method. Its key move is to rewrite one middle term as two carefully chosen terms, then factor the resulting four terms by grouping.

Why these quadratics are harder

For a monic trinomial such as

x^{2} + 7 x + 12,

you seek two integers that multiply to $12$ and add to $7$ . That immediately gives

(x + 3) (x + 4) .

But with a non- $1$ leading coefficient, the multiplication behind the factorization is less visible. For instance,

6 x^{2} - x - 15

does not tell us directly which two numbers should multiply to $- 15$ , because the $6 x^{2}$ must itself arise from multiplying the $x$ -terms in two binomials.

Suppose the desired factorization has the form

(p x + q) (r x + s) .

Expanding gives

(p x + q) (r x + s) = p r x^{2} + (p s + q r) x + q s .

Matching this with

a x^{2} + b x + c

shows that

a = p r, b = p s + q r, c = q s .

Now focus on the two contributions to the middle term:

m = p s, n = q r .

They have two crucial properties:

m + n = b

and

mn = (p s) (q r) = (p r) (q s) = a c .

That is the logic of the $a c$ method:

Find two integers whose product is $a c$ and whose sum is $b$ .

Those numbers allow us to split $b x$ into $m x + n x$ . The expression becomes factorable by grouping.

See the method once, visually

Factoring trinomials with a non-1 leading coefficient by grouping

Watch “Factoring trinomials with a non-1 leading coefficient by grouping” from Khan Academy. It gives a compact visual demonstration of the exact logic behind the ac method, including the sign reasoning needed when ac is negative.

Start with the target conditions: the two replacement coefficients must multiply to ac and add to b. Then watch the factor search, paying close attention to why a negative product requires opposite signs. Finish with the grouping step, where the split middle term produces a repeated binomial factor.

The video factors

4 y^{2} + 4 y - 15.

Here,

a c = 4 (- 15) = - 60.

The needed pair is $10$ and $- 6$ , since

10 (- 6) = - 60

and

10 + (- 6) = 4.

So the middle term becomes $10 y - 6 y$ , after which grouping exposes the factors.

The dependable $a c$ routine

For a trinomial

a x^{2} + b x + c,

use this sequence.

Factor out a GCF first.
This is non-negotiable: it simplifies the remaining work and ensures your final answer is fully factored.
Compute $a c$ .
Multiply the leading coefficient $a$ by the constant term $c$ .
Find integers $m$ and $n$ satisfying
$mn = a c$
and
$m + n = b .$
Split the middle term.
Replace $b x$ with $m x + n x$ .
Factor by grouping.
Group the first two and final two terms. Factor the GCF from each group, then factor out the repeated binomial.
Check by expanding.
Multiply the binomial factors to verify the original trinomial is reproduced exactly.

The two conditions on $m$ and $n$ must not be mixed up:

Condition	What it must equal
Product $mn$	$a c$
Sum $m + n$	$b$

A complete example: $6 x^{2} - x - 15$

Consider

6 x^{2} - x - 15.

There is no GCF shared by all three terms. Identify the coefficients:

a = 6, b = - 1, c = - 15.

First calculate:

a c = 6 (- 15) = - 90.

We need two integers with product $- 90$ and sum $- 1$ . The correct pair is

- 10 and 9,

because

(- 10) (9) = - 90

and

- 10 + 9 = - 1.

Split the middle term:

6 x^{2} - x - 15 = 6 x^{2} - 10 x + 9 x - 15.

Now group:

(6 x^{2} - 10 x) + (9 x - 15) .

Factor the GCF from each group:

2 x (3 x - 5) + 3 (3 x - 5) .

Both terms contain $(3 x - 5)$ , so factor it out:

(3 x - 5) (2 x + 3) .

Thus,

6 x^{2} - x - 15 = (3 x - 5) (2 x + 3) .

This worked visual shows $6x^2-x-15$ rewritten as $6x^2-10x+9x-15$, then factored by grouping into $(3x-5)(2x+3)$. The pair $-10$ and $9$ has product $-90=6(-15)$ and sum $-1$, which is exactly why the middle term can be split this way.

A quick expansion checks the result:

(3 x - 5) (2 x + 3) = 6 x^{2} + 9 x - 10 x - 15 = 6 x^{2} - x - 15.

Sign reasoning without guesswork

The sign of $a c$ predicts what signs the two split coefficients must have.

When $a c > 0$

The two integers have the same sign.

If $b > 0$ , both are positive.
If $b < 0$ , both are negative.

For example, if

a c = 28 and b = - 11,

the pair must consist of two negative integers:

(- 7) (- 4) = 28, - 7 + (- 4) = - 11.

When $a c < 0$

The two integers have opposite signs.

The integer with the greater absolute value must have the same sign as $b$ . For example, if

a c = - 90 and b = - 1,

the negative number must have slightly greater magnitude:

- 10 + 9 = - 1.

This sign check narrows the search substantially before you list factor pairs.

Example with a GCF first

Now factor

10 y^{2} - 55 y + 70.

The GCF is $5$ , so begin there:

10 y^{2} - 55 y + 70 = 5 (2 y^{2} - 11 y + 14) .

Only now apply the $a c$ method to the trinomial inside the parentheses. Here,

a = 2, b = - 11, c = 14.

Thus,

a c = 2 (14) = 28.

We need two integers that multiply to $28$ and add to $- 11$ :

- 7 and - 4.

Rewrite the middle term:

5 (2 y^{2} - 11 y + 14) = 5 (2 y^{2} - 7 y - 4 y + 14) .

Group the terms:

5 [(2 y^{2} - 7 y) + (- 4 y + 14)] .

Factor each group:

5 [y (2 y - 7) - 2 (2 y - 7)] .

Now the repeated binomial is visible:

5 (y - 2) (2 y - 7) .

Therefore,

10 y^{2} - 55 y + 70 = 5 (y - 2) (2 y - 7) .

The outer $5$ is essential. Writing only $(y - 2) (2 y - 7)$ would factor the reduced trinomial, not the original one.

What “over the integers” means

This lesson concerns factorizations whose factors have integer coefficients. For example,

4 x^{2} + 8 x - 5

factors over the integers because

4 x^{2} + 8 x - 5 = (2 x + 5) (2 x - 1) .

Indeed,

a c = 4 (- 5) = - 20,

and the required pair is $10$ and $- 2$ :

10 (- 2) = - 20, 10 + (- 2) = 8.

By contrast, consider

2 x^{2} + x + 2.

Here,

a c = 2 (2) = 4.

The integer factor pairs of $4$ are $1$ and $4$ , or $2$ and $2$ , with signs adjusted together because the product is positive. Their possible sums are

5, - 5, 4, or - 4,

but never $1$ . Therefore, there are no integers $m$ and $n$ satisfying the $a c$ conditions, so

2 x^{2} + x + 2

does not factor into binomials with integer coefficients.

This is not a failure of technique. It is a legitimate conclusion: some quadratic trinomials are not factorable over the integers.

Grouping is the bridge, not an extra mystery

After splitting the middle term, your aim is to create a common binomial. In the earlier example,

6 x^{2} - 10 x + 9 x - 15,

the first pair gave

2 x (3 x - 5),

and the second pair gave

3 (3 x - 5) .

The repeated factor $(3 x - 5)$ is what makes the factorization possible.

Sometimes one group begins with negative terms. Then factor out a negative GCF if necessary to make the binomials match. For example,

4 x^{2} + 10 x - 2 x - 5

groups as

(4 x^{2} + 10 x) + (- 2 x - 5) .

Factoring each group gives

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

The $- 1$ is not cosmetic: it preserves the signs and reveals the common binomial. The result is

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

At this stage, you are using grouping in a focused way: the $a c$ method has deliberately manufactured the four-term expression that grouping needs. A later lesson will extend grouping to more general four-term polynomials.

Common errors to catch early

Error	Why it causes trouble	Correction
Skipping the initial GCF	The answer may be incomplete, and the numbers in the $a c$ step become unnecessarily large.	Always scan for a GCF first.
Finding factors of $c$ instead of $a c$	This only works reliably when the leading coefficient is $1$ .	Use the product $a \cdot c$ .
Finding a pair with the right product but wrong sum	The split terms will not recombine to the original middle term.	Verify both conditions before rewriting.
Losing a sign while splitting $b x$	The rewritten polynomial is no longer equivalent to the original.	Add the two proposed coefficients explicitly.
Factoring a positive GCF when a negative one is needed	The grouped binomials may look like opposites rather than matches.	Factor out $- 1$ , or another negative GCF, from one group.
Stopping at a partly factored expression	A shared binomial may still be available.	Look for the repeated parenthetical factor after grouping.

A final expansion is the fastest universal check. If the factors expand to the original three terms with the correct coefficients and signs, the factorization is correct.

Key takeaways

To factor a quadratic trinomial with a leading coefficient other than $1$ :

factor out any GCF first;
identify $a$ , $b$ , and $c$ in $a x^{2} + b x + c$ ;
find integers $m$ and $n$ such that

mn = a c

and

m + n = b;

replace $b x$ with $m x + n x$ ;
factor the resulting four terms by grouping;
expand to check the result;
if no integer pair meets both conditions, the trinomial does not factor over the integers.