Quadratics & their graphs · Lesson 2 of 6

Solving by factorising

Turn ax² + bx + c into two brackets, read off the roots, build the equation from its roots, and find the sum and product of the roots.

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Factorising is the quickest way to solve a quadratic, when it works. It rests on one simple fact.

Why brackets give the answers

If two numbers multiply to give 0, at least one of them must be 0. There’s no other way to get 0 from a multiplication.

So if (x−3)(x+2)=0(x - 3)(x + 2) = 0, then either x−3=0x - 3 = 0 or x+2=0x + 2 = 0. That gives x=3x = 3 or x=−2x = -2.

Finding the brackets

Multiply out (x+3)(x+4)(x + 3)(x + 4) and you get x2+7x+12x^2 + 7x + 12. Notice that 3×4=123 \times 4 = 12 and 3+4=73 + 4 = 7. Factorising runs this backwards:

  • To factorise x2+bx+cx^2 + bx + c, find two numbers that multiply to cc and add to bb.
  • When there’s a number in front of x2x^2, as in ax2+bx+cax^2 + bx + c, find two numbers that multiply to acac and add to bb. Use them to split the middle term into two, then take out common factors in pairs.

Try it. Pick the right pair; a wrong pick tells you which part failed.

Find the two numbersPick a pair

Solve x² + 7x + 12 = 0

Look for two numbers that multiply to c = 12 and add to b = 7. The product is positive, so both numbers have the same sign as b (positive).

A past question, step by step

Worked example · WAEC 2025

WAEC 2025 · Paper 1 · Q16

Solve 3x2+25x−18=03x^2 + 25x - 18 = 0.

  1. Multiply a by c

    There’s a 3 in front of x2x^2, so we need two numbers that multiply to ac=3×(−18)=−54ac = 3 \times (-18) = -54 and add to b=25b = 25.

    Think first. Which two numbers multiply to −54-54 and add to 2525?

  2. Find the pair

    27×(−2)=−5427 \times (-2) = -54 and 27+(−2)=2527 + (-2) = 25. So split 25x25x into 27x−2x27x - 2x:

    3x2+27x−2x−18=03x^2 + 27x - 2x - 18 = 0
  3. Take out common factors in pairs

    3x(x+9)−2(x+9)=03x(x + 9) - 2(x + 9) = 0(3x−2)(x+9)=0(3x - 2)(x + 9) = 0
  4. Read off the roots

    3x−2=03x - 2 = 0 gives x=23x = \frac23, and x+9=0x + 9 = 0 gives x=−9x = -9. The answer is B.

Going backwards: from roots to the equation

Exam questions often give you the roots and ask for the equation. Turn each root into a bracket, then multiply out.

  • A root x=3x = 3 comes from the bracket (x−3)(x - 3).
  • A fraction root such as x=−35x = -\frac35: clear the fraction first. 5x=−35x = -3, so 5x+3=05x + 3 = 0, and the bracket is (5x+3)(5x + 3).

So the equation with roots 44 and −35-\frac35 is (x−4)(5x+3)=0(x - 4)(5x + 3) = 0, which is 5x2−17x−12=05x^2 - 17x - 12 = 0.

More: from the roots to the equation

The sum and product of the roots

If α\alpha and β\beta are the roots of ax2+bx+c=0ax^2 + bx + c = 0, you can find their sum and product without solving:

α+β=−baαβ=ca\alpha + \beta = -\frac ba \qquad \alpha\beta = \frac ca

For example, for 2x2−7x+3=02x^2 - 7x + 3 = 0 the roots add to 72\frac72 and multiply to 32\frac32. Other expressions are built from these two: 1α+1β=α+βαβ\frac1\alpha + \frac1\beta = \frac{\alpha + \beta}{\alpha\beta}.

ax² + bx + c = 0, roots α and β
α + β = −b⁄a αβ = c⁄a
The equation is x² − (sum)x + (product) = 0
Sum and product of the rootsSum −b/a, product c/a

More: the sum and product of the roots

Your turn

WAEC 2023 · Paper 1 · Q9

Find the roots of the equation 3m2−2m−65=03m^2 - 2m - 65 = 0.

Worked solution (try it first)
  1. Find two numbers with product 3×(−65)=−1953 \times (-65) = -195 and sum −2-2: they are −15-15 and 13.
  2. Split and group: 3m2−15m+13m−65=3m(m−5)+13(m−5)3m^2 - 15m + 13m - 65 = 3m(m - 5) + 13(m - 5)
    =(3m+13)(m−5)= (3m + 13)(m - 5).
  3. So m=5m = 5 or m=−133m = -\frac{13}{3}, option C.

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