Factoring quadratics when a is not 1

TL;DR:

• For ax² + bx + c, find two numbers that multiply to ac and add to b.

• Use them to split the middle term, then factor by grouping.

• Always check by expanding. If no whole-number pair exists, use the quadratic formula.

Factoring x² + 5x + 6 is a guessing game with small numbers. Factoring 6x² + 11x − 10 feels harder because the leading 6 gives many more combinations. The ac method turns the guess into a search with a clear target.

What factoring is undoing

Multiplying (3x − 2)(2x + 5) gives four products: 6x², 15x, −4x and −10. The two middle terms, 15x and −4x, combine into the single 11x you see in the final answer. Factoring runs this backwards: it splits the 11x into the two pieces that came from the multiplication.

Those two pieces are the clue. Their product is 15x × (−4x) = −60x², which equals 6x² × (−10), the first coefficient times the last. Their sum is the middle coefficient, 11.

Worked example: 6x² + 11x − 10

1. Multiply the outer coefficients: a × c = 6 × (−10) = −60.

2. Find two numbers that multiply to −60 and add to 11. The product is negative, so one is positive and one is negative, and the larger in size is positive. Trying pairs: 15 and −4 give product −60 and sum 11.

3. Split the middle term: 6x² + 15x − 4x − 10.

4. Group the first two and the last two: 3x(2x + 5) − 2(2x + 5). Factor out the common (2x + 5).

5. Result: (3x − 2)(2x + 5).

Check by expanding

Multiply the answer back out: 3x · 2x = 6x², 3x · 5 = 15x, −2 · 2x = −4x, −2 · 5 = −10. Adding gives 6x² + 11x − 10. It matches.

You can also check the roots. Setting each factor to zero gives x = 2/3 and x = −5/2. Test x = 2/3: 6 · 4/9 = 8/3, and 11 · 2/3 = 22/3, so the sum is 30/3 = 10, and 10 − 10 = 0. Test x = −5/2: 6 · 25/4 = 37.5, 11 · (−2.5) = −27.5, and 37.5 − 27.5 − 10 = 0. Both work.

How to find the pair of numbers faster

• List the factor pairs of the product ac and look for the one that adds to b.

• Use the signs first: if ac is negative, the pair has opposite signs; if ac is positive, both share the sign of b.

• If the middle coefficient is small compared with the product, expect the two numbers to be far apart in size.

For −60, the pairs are 1, −60, 2, −30, 3, −20, 4, −15, 5, −12, 6, −10 and so on, in both sign arrangements. Only 15 and −4 add to 11.

Try one yourself: 4x² − 4x − 15

Here ac = 4 × (−15) = −60 again, but now the middle coefficient is −4. Look for a pair with product −60 and sum −4: 6 and −10 work, since 6 − 10 = −4. Split the middle: 4x² + 6x − 10x − 15. Group: 2x(2x + 3) − 5(2x + 3). Factor: (2x − 5)(2x + 3).

Check by expanding: 4x² + 6x − 10x − 15 = 4x² − 4x − 15. Correct. The roots are x = 5/2 and x = −3/2.

Why splitting the middle term works

Grouping only succeeds when both groups share the same bracket. That happens exactly when the two split pieces multiply to ac. In the first example the bracket (2x + 5) appeared in both groups, which is the signal that you chose the right pair. If the two brackets come out different, the pair is wrong or an arithmetic slip has crept in, so go back one step instead of forcing it.

When to switch to the formula

Take 2x² + 3x − 7. Here ac = −14, and no pair of whole numbers multiplies to −14 and adds to 3. The discriminant confirms it: b² − 4ac = 9 + 56 = 65, which is not a perfect square. If the discriminant is not a perfect square, the quadratic does not factor over the integers, so go straight to the quadratic formula. If it is a perfect square, factoring will work, and you can use the discriminant as a quick test before searching.

Common mistakes

• Forgetting to factor out a common number first, such as the 2 in 4x² + 10x − 6.

• Getting the signs wrong when pulling a negative out of the second group.

• Stopping before the final factoring step, leaving 3x(2x + 5) − 2(2x + 5) as the answer.

• Skipping the expansion check.

A quicker way to see which pair to try

Before listing every pair, think about the end of the process. The last term of the answer, −10, came from multiplying the two constants in the brackets, here −2 and 5. The first term, 6x², came from multiplying 3x and 2x. So the constants must be a factor pair of 10 and the x-coefficients a factor pair of 6. That gives a short list: for 6 you have 1 × 6 or 2 × 3, and for 10 you have 1 × 10 or 2 × 5.

Trying them in the brackets and checking the middle term is the "box" or "trial" method. It is the same search as the ac method, only organised differently. If you prefer one, stick with it, and use the other as a cross-check when an answer looks suspicious.

Where you will use it next

Factored form shows the roots at a glance, which is useful for graphing and for simplifying fractions like (6x² + 11x − 10)/(2x + 5), which equals 3x − 2 when x is not −5/2. A related method, completing the square, works on every quadratic, including those that do not factor.

Frequently asked questions

Does the order of the two factors matter?

No. Multiplication is commutative, so (3x − 2)(2x + 5) and (2x + 5)(3x − 2) are the same.

What if a is negative?

Factor out −1 first. Then −2x² + 5x + 3 becomes −(2x² − 5x − 3), and you factor the bracket as usual.