How to Factor Quadratic Equations
Factoring turns a quadratic into a product of two brackets, and a product equals zero only when one of its factors is zero. That single fact is why factoring solves equations at all.
When the leading coefficient is 1
- For x^2 + bx + c, find two numbers that multiply to c and add to b.
- Those two numbers go straight into the brackets: (x + p)(x + q).
- Set each bracket to zero and solve.
When it is not 1
For ax^2 + bx + c, find two numbers that multiply to a times c and add to b. Split the middle term using those two numbers, then factor the four terms in pairs. This method always works and never needs guessing.
Patterns worth recognising
- Difference of squares: x^2 - 36 = (x - 6)(x + 6).
- Perfect square: x^2 + 10x + 25 = (x + 5)^2.
- Common factor first: 2x^2 + 8x = 2x(x + 4), before anything else.
Worked examples
Example 1
Question
Factor and solve: x^2 + 7x + 12 = 0
Answer
x = -3 or x = -4
Steps
- 1Leading coefficient is 1, so look for two numbers multiplying to 12 and adding to 7.
- 2Those are 3 and 4.
- 3So x^2 + 7x + 12 = (x + 3)(x + 4).
- 4Set each factor to zero: x + 3 = 0 gives x = -3, and x + 4 = 0 gives x = -4.
- 5Check: (-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0.
Example 2
Question
Factor and solve: x^2 - 5x - 14 = 0
Answer
x = 7 or x = -2
Steps
- 1Two numbers multiplying to -14 and adding to -5.
- 2The product is negative, so the numbers have opposite signs: -7 and 2.
- 3So x^2 - 5x - 14 = (x - 7)(x + 2).
- 4Solutions: x = 7 and x = -2.
- 5Check: 7^2 - 5(7) - 14 = 49 - 35 - 14 = 0.
A negative constant term always means the two numbers have opposite signs. That halves the list you have to search.
Example 3
Question
Factor and solve: 2x^2 + 5x - 3 = 0
Answer
x = 1/2 or x = -3
Steps
- 1Leading coefficient is not 1, so use a times c: 2 x (-3) = -6.
- 2Find two numbers multiplying to -6 and adding to 5: they are 6 and -1.
- 3Split the middle term: 2x^2 + 6x - x - 3 = 0.
- 4Factor in pairs: 2x(x + 3) - 1(x + 3) = 0.
- 5Factor out the common bracket: (2x - 1)(x + 3) = 0.
- 6Solutions: 2x - 1 = 0 gives x = 1/2, and x + 3 = 0 gives x = -3.
Common mistakes
Factoring correctly and then forgetting to set each bracket to zero.
The factored form is not the answer to an equation. (x + 3)(x + 4) = 0 still needs solving.
Looking for numbers that add to c and multiply to b.
Multiply to c, add to b. Getting them the wrong way round is the single most common slip.
Missing a common factor: factoring 2x^2 + 10x + 12 without taking out the 2 first.
Always check for a common factor before anything else. It makes everything after it easier.
Assuming every quadratic factors neatly.
Check the discriminant, b^2 - 4ac. If it is not a perfect square, use the quadratic formula instead of hunting for factors that are not there.
Practice it
Work these out first, then open the answer. Stuck halfway? Select the problem and ask Solvecoon for the step you are missing.
1. Factor and solve: x^2 + 9x + 20 = 0
Show answer
(x + 4)(x + 5) = 0, so x = -4 or x = -5.
2. Factor and solve: x^2 - 36 = 0
Show answer
Difference of squares: (x - 6)(x + 6) = 0, so x = 6 or x = -6.
3. Factor and solve: 3x^2 - 10x + 8 = 0
Show answer
a x c = 24, and -6 and -4 add to -10. This gives (3x - 4)(x - 2) = 0, so x = 4/3 or x = 2.
Questions people ask
- How do I know whether a quadratic factors?
- Work out the discriminant, b^2 - 4ac. If it is a perfect square, the quadratic factors over the integers. If it is not, do not waste time hunting.
- Is factoring faster than the quadratic formula?
- When it works, yes, considerably. When it does not, it is slower than useless because you spend the time and still have to use the formula.
- Why does setting each factor to zero work?
- Because the only way a product of two numbers is zero is if at least one of them is zero. That is the zero product property, and it is the whole reason we factor.
- What if there is no constant term?
- Factor out the x: x^2 - 5x = x(x - 5), so x = 0 or x = 5. Dividing both sides by x instead loses the x = 0 solution.
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