The product-sum method for factoring ax² + bx + c

By Marius, private math tutor in Montreal · 2026-10-05

To factor $a x^{2} + b x + c$ by the product-sum method, find two numbers whose product is $a \times c$ and whose sum is $b$ , use them to split the middle term $b x$ into two terms, and factor the four terms by grouping. It works for every trinomial that factors over the integers, including the ones where $a \neq 1$ , and when no such pair exists, it tells you so.

Why the two numbers multiply to a × c

Multiply out any two binomials and watch where the middle term comes from:

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

The leading coefficient is $a = p r$ and the constant is $c = q s$ .
The two middle pieces, $p s$ and $q r$ , add up to $b$ .
Their product is $p s \times q r = (p r) (q s)$ , which is $a \times c$ .

So every trinomial that factors hides two numbers with product $a c$ and sum $b$ : they are the two middle terms from before they were combined. The method finds them and splits the middle term back into those two pieces. The four terms then factor by grouping, because that is exactly how they were produced. When $a = 1$ , the product $a c$ is just $c$ , which is the familiar search for $x^{2} + b x + c$ .

The method in five steps

Take out any common factor first. Smaller coefficients mean fewer pairs to try.
Compute $a \times c$ , and let the signs narrow the search. If $a c > 0$ , both numbers have the sign of $b$ . If $a c < 0$ , they have opposite signs, and the one farther from zero takes the sign of $b$ .
List the factor pairs of the size of $a c$ until one gives the right sum (or, for opposite signs, the right difference).
Split $b x$ into two terms using the pair.
Group the terms in pairs, take the common factor out of each pair, then take out the common bracket. Check by expanding.

Worked example 1

Factor $20 x^{2} - 3 x - 2$ .

Common factor: none; 20, 3 and 2 share no factor.
Product and sum: $a c = 20 \times (- 2) = - 40$ and $b = - 3$ . The product is negative, so the signs differ; the sum is negative, so the number farther from zero is the negative one.
Pairs of 40: $1 \times 40$ , $2 \times 20$ , $4 \times 10$ , $5 \times 8$ . The pair that differs by 3 is $5$ and $8$ , so the numbers are $5$ and $- 8$ : $5 \times (- 8) = - 40$ and $5 + (- 8) = - 3$ .
Split: $20 x^{2} - 3 x - 2 = 20 x^{2} + 5 x - 8 x - 2$
Group: $= 5 x (4 x + 1) - 2 (4 x + 1)$ $= (4 x + 1) (5 x - 2)$
Check: $(4 x + 1) (5 x - 2) = 20 x^{2} - 8 x + 5 x - 2 = 20 x^{2} - 3 x - 2$ .

In the second pair, take out $- 2$ , not $2$ : choose the sign that leaves the same bracket as the first pair, $4 x + 1$ . The order of the split does not matter either. Writing $20 x^{2} - 8 x + 5 x - 2$ groups as $4 x (5 x - 2) + 1 (5 x - 2)$ and gives the same two factors.

Worked example 2: the common factor comes first

Factor $12 x^{2} + 34 x + 14$ .

Every coefficient is even: $12 x^{2} + 34 x + 14 = 2 (6 x^{2} + 17 x + 7)$ .
For $6 x^{2} + 17 x + 7$ : $a c = 42$ and $b = 17$ , both positive, so both numbers are positive. Pairs of 42: $1 \times 42$ , $2 \times 21$ , $3 \times 14$ , $6 \times 7$ . The sum 17 comes from $3$ and $14$ .
Split and group: $6 x^{2} + 3 x + 14 x + 7 = 3 x (2 x + 1) + 7 (2 x + 1) = (2 x + 1) (3 x + 7)$
Bring the 2 back: $12 x^{2} + 34 x + 14 = 2 (2 x + 1) (3 x + 7)$ .

Skipping the first step still works, but it costs you. The product becomes $12 \times 14 = 168$ , which has twice as many factor pairs to search, and the 2 still has to be pulled out of a bracket at the end: the split $6 x + 28 x$ leads to $(2 x + 1) (6 x + 14)$ .

The most common mistake. When $a \neq 1$ , the two numbers multiply to $a \times c$ , not to $c$ . The habit carried over from trinomials like $x^{2} + b x + c$ is to look for a product of $c$ . With $a \neq 1$ that search either finds nothing, and the student wrongly decides the trinomial cannot be factored, or finds a pair whose split will not group. Make $a c$ the first thing you write on every product-sum question.

When no pair exists

If you have listed every factor pair of $a c$ and none has the right sum, the trinomial does not factor over the integers, and that is a legitimate answer. Take $2 x^{2} + 5 x + 4$ : $a c = 8$ , and the pairs $1 \times 8$ and $2 \times 4$ add to 9 and 6, never 5 (the negative pairs give negative sums).

There is a quick confirmation. The two numbers you are looking for are the solutions of $t^{2} - b t + a c = 0$ . When $a$ , $b$ and $c$ are integers, as they are in every product-sum question, those solutions are integers exactly when the discriminant $b^{2} - 4 a c$ is a perfect square ( $0$ , $1$ , $4$ , $9$ , $16$ , and so on). For $2 x^{2} + 5 x + 4$ it is $25 - 32 = - 7$ , so no pair exists.

Optional, and beyond what Secondary 4 asks. When $a c$ is large and the discriminant is a perfect square, the same fact hands you the pair directly:

\frac{b + \sqrt{b^{2} - 4 a c}}{2} and \frac{b - \sqrt{b^{2} - 4 a c}}{2}

In worked example 1, $b^{2} - 4 a c = 9 + 160 = 169 = 13^{2}$ , and half of $- 3 + 13$ and of $- 3 - 13$ gives $5$ and $- 8$ .

A trinomial that will not factor still has a standard form, and completing the square always gets you there. One that does factor is now in factored form, and what each form of a quadratic function tells you explains what to read off it.

Check your understanding

No calculator. Check every factorization by expanding it.

Find two numbers whose product is $- 36$ and whose sum is $5$ .
Factor $3 x^{2} + 10 x + 8$ .
Factor $9 x^{2} + 3 x - 2$ .
Factor $12 x^{2} - 17 x + 6$ .
Factor completely: $6 x^{3} + 3 x^{2} - 9 x$ .
Explain why, when $a \times c$ is positive and $b$ is negative, both numbers in the product-sum method must be negative.
Léa wants to factor $3 x^{2} - 10 x - 8$ . She looks for two numbers with product $- 8$ and sum $- 10$ , finds none, and concludes that the trinomial cannot be factored. Find her error, then factor the trinomial.
True or false: because no two integers have product $8$ and sum $7$ , the function $f (x) = 2 x^{2} + 7 x + 4$ has no zeroes. Explain.

Show the answers

$9$ and $- 4$ : $9 \times (- 4) = - 36$ and $9 + (- 4) = 5$ . The product is negative, so the signs differ, and the sum is positive, so the larger number is the positive one.
$(3 x + 4) (x + 2)$ . Here $a c = 24$ with sum $10$ , so the numbers are $4$ and $6$ : $3 x^{2} + 4 x + 6 x + 8 = x (3 x + 4) + 2 (3 x + 4)$ .
$(3 x + 2) (3 x - 1)$ . Here $a c = - 18$ with sum $3$ , so the numbers are $6$ and $- 3$ : $9 x^{2} + 6 x - 3 x - 2 = 3 x (3 x + 2) - 1 (3 x + 2)$ . The second pair gives up $- 1$ so that the brackets match.
$(3 x - 2) (4 x - 3)$ . Here $a c = 72$ with sum $- 17$ , so both numbers are negative: $- 8$ and $- 9$ . Then $12 x^{2} - 8 x - 9 x + 6 = 4 x (3 x - 2) - 3 (3 x - 2)$ .
$3 x (2 x + 3) (x - 1)$ . The common factor $3 x$ comes out first, leaving $2 x^{2} + x - 3$ , with $a c = - 6$ and sum $1$ : the numbers $3$ and $- 2$ give $2 x^{2} + 3 x - 2 x - 3 = x (2 x + 3) - 1 (2 x + 3)$ .
A positive product means the two numbers have the same sign. Two positive numbers have a positive sum, so if the sum $b$ is negative, both numbers have to be negative.
Her product is wrong: with $a = 3$ , the two numbers must multiply to $3 \times (- 8) = - 24$ , not $- 8$ . With sum $- 10$ they are $- 12$ and $2$ , and $3 x^{2} + 2 x - 12 x - 8 = x (3 x + 2) - 4 (3 x + 2) = (3 x + 2) (x - 4)$ .
False. No integer pair only means that $f$ has no factored form with integer coefficients. Its discriminant is $7^{2} - 4 (2) (4) = 17 > 0$ , so $f$ has two real zeroes, $\frac{- 7 \pm \sqrt{17}}{4}$ ; they are simply not rational. This is the perfect-square test from the section above: $17$ is positive, so there are real zeroes, but it is not a perfect square, so there is no integer pair.

Keep practising

The free Secondary 4 Algebraic Expressions worksheet has a product-sum question in a set that mixes every factoring technique, which is the real test: deciding which one a trinomial needs. The Quadratic Functions worksheet then uses factoring to find the zeroes of a function.

Want one-on-one help with this? I offer Secondary 4 and 5 math tutoring in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.