Solving Quadratics by Factorising
If a product is zero, one of its factors must be zero. So once \(x^2 - x - 12 = 0\) is factorised as \((x - 4)(x + 3) = 0\), the solutions are \(x = 4\) and \(x = -3\). This NSW Year 9 Advanced maths subtopic solves quadratic equations this way (NESA outcome MA5‑EQU‑P‑01).
You will solve monic and non-monic equations, rearrange equations so one side is zero, and clear fractions when \(x\) is in a denominator. You will explain why dividing by \(x\) loses a solution, find a missing coefficient from a known solution, and solve problems about consecutive integers, a garden, a ball's height, a right-angled triangle and a path around a lawn. It is part of Year 9 Advanced Quadratics.
Every question with a fully worked solution.
- Solving Quadratics by Factorising - Video - Solving Quadratic Equations by Factoring Watch
Theory
If two numbers multiply to give zero, at least one of them must be zero. This null factor law turns a factorised quadratic into two simple linear equations, so most quadratics have two solutions. This Year 9 Advanced maths page shows how to rearrange, factorise and solve, and how to use the method in problems. It leads on to Advanced and Extension maths.
A quadratic equation can be written as \(ax^2 + bx + c = 0\) with \(a \ne 0\). Its solutions (or roots) are the values of \(x\) that make it true.
The null factor law: if \(A \times B = 0\), then \(A = 0\) or \(B = 0\). It only works when one side of the equation is zero.
A quadratic equation has two solutions, one repeated solution (when it is a perfect square, such as \((x - 3)^2 = 0\)), or no real solutions.
The solutions of \(ax^2 + bx + c = 0\) are the \(x\)-intercepts of the parabola \(y = ax^2 + bx + c\).
In a problem, a solution that makes no sense, such as a negative length, is rejected.
| Equation | Factorised | Solutions |
|---|---|---|
| \(x^2 - 8x + 12 = 0\) | \((x - 2)(x - 6) = 0\) | \(x = 2\) or \(6\) |
| \(x^2 - 49 = 0\) | \((x - 7)(x + 7) = 0\) | \(x = \pm 7\) |
| \(x^2 + 7x = 0\) | \(x(x + 7) = 0\) | \(x = 0\) or \(-7\) |
| \(x^2 - 6x + 9 = 0\) | \((x - 3)^2 = 0\) | \(x = 3\) (repeated) |
How to solve by factorising
- Make one side zero. Expand any brackets, clear fractions (multiply by the denominator, noting it cannot be zero), and move every term to one side.
- Simplify: divide through by a common number if you can, but never divide by \(x\).
- Factorise completely: HCF, product and sum, the ac method, or a special product.
- Set each factor to zero and solve each linear equation.
- Check by substituting into the original equation, and reject any answer that does not fit the context.
Problems
- Name the unknown with a letter and write the other quantities in terms of it (consecutive numbers \(n\) and \(n + 1\); a length \(5\) more is \(w + 5\)).
- Form the equation from a formula: area, Pythagoras' theorem, or a given rule such as \(h = 20t - 5t^2\).
- A known solution: substitute it to find an unknown coefficient, then solve the full equation.
- \(x^4\) and \(x^2\): let \(u = x^2\), solve for \(u\), then take square roots, \(\pm\).
- Answer the question in words, with units.
(i) \(x^2 - 3x - 28 = 0\)
(ii) \(9x^2 = 25\)
(iii) \(5x^2 = 15x\)
(iv) \(x^2 + 10x + 25 = 0\)
| \(x^2 - 3x - 28\) | \(=\) | \(0\) |
| \((x - 7)(x + 4)\) | \(=\) | \(0\) |
| \(x - 7 = 0\) | \(\text{or}\) | \(x + 4 = 0\) |
| \(\boldsymbol{x = 7}\) | \(\text{or}\) | \(\boldsymbol{x = -4}\) |
(ii) Move \(25\) across: a difference of two squares.
| \(9x^2 - 25\) | \(=\) | \(0\) |
| \((3x - 5)(3x + 5)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{5}{3}} \ \text{or}\ {\boldsymbol{-\dfrac{5}{3}}}\) |
(iii) Do not divide by \(x\): that loses the solution \(x = 0\). Bring everything to one side and take out the HCF.
| \(5x^2 - 15x\) | \(=\) | \(0\) |
| \(5x(x - 3)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{0} \ \text{or}\ \boldsymbol{3}\) |
(iv) A perfect square gives one repeated solution.
| \((x + 5)^2\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{-5}\) |
(i) \((x + 2)(x - 5) = 8\)
(ii) \(2x^2 + 5x = 3\)
(iii) \(x - \dfrac{10}{x} = 3\)
(i) The right side is \(8\), not \(0\), so the factors cannot be set to zero yet. Expand and rearrange.
| \(x^2 - 3x - 10\) | \(=\) | \(8\) |
| \(x^2 - 3x - 18\) | \(=\) | \(0\) |
| \((x - 6)(x + 3)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{6} \ \text{or}\ {\boldsymbol{-3}}\) |
(ii) Subtract \(3\), then factorise the non-monic trinomial (\(ac = -6\): \(6\) and \(-1\)).
| \(2x^2 + 5x - 3\) | \(=\) | \(0\) |
| \((2x - 1)(x + 3)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{1}{2}} \ \text{or}\ {\boldsymbol{-3}}\) |
(iii) \(x \ne 0\), so multiply every term by \(x\).
| \(x^2 - 10\) | \(=\) | \(3x\) |
| \(x^2 - 3x - 10\) | \(=\) | \(0\) |
| \((x - 5)(x + 2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{5} \ \text{or}\ {\boldsymbol{-2}}\) |
Check (iii) with \(x = -2\): \(-2 - \dfrac{10}{-2} = -2 + 5 = 3\ \checkmark\)
(i) \(x = -2\) is a solution of \(2x^2 + kx - 6 = 0\). Find \(k\) and the other solution.
(ii) Solve \(x^4 - 17x^2 + 16 = 0\).
(i) A solution makes the equation true, so substitute it.
| \(2(-2)^2 + k(-2) - 6\) | \(=\) | \(0\) |
| \(8 - 2k - 6\) | \(=\) | \(0\) |
| \(k\) | \(=\) | \(\boldsymbol{1}\) |
| \(2x^2 + x - 6\) | \(=\) | \(0\) |
| \((2x - 3)(x + 2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\dfrac{3}{2} \ \text{or}\ {-2}\) |
The other solution is \(\boldsymbol{x = \dfrac{3}{2}}\).
(ii) Let \(u = x^2\). Product \(16\), sum \(-17\): \(-1\) and \(-16\).
| \(u^2 - 17u + 16\) | \(=\) | \(0\) |
| \((u - 1)(u - 16)\) | \(=\) | \(0\) |
| \(x^2 = 1\) | \(\text{or}\) | \(x^2 = 16\) |
| \(x\) | \(=\) | \(\boldsymbol{\pm 1}\ \text{or}\ \boldsymbol{\pm 4}\) |
Each value of \(x^2\) gives two values of \(x\), so there are four solutions.
Area \(= \dfrac{1}{2} \times\) base \(\times\) height.
| \(\dfrac{1}{2}h(h + 4)\) | \(=\) | \(30\) |
| \(h^2 + 4h\) | \(=\) | \(60\) |
| \(h^2 + 4h - 60\) | \(=\) | \(0\) |
| \((h + 10)(h - 6)\) | \(=\) | \(0\) |
| \(h\) | \(=\) | \({-10} \ \text{or}\ 6\) |
A length cannot be negative, so reject \(h = -10\).
| \(\text{height}\) | \(=\) | \(\boldsymbol{6}~\text{cm}\) |
| \(\text{base}\) | \(=\) | \(6 + 4 = \boldsymbol{10}~\text{cm}\) |
Check: \(\dfrac{1}{2} \times 10 \times 6 = 30~\text{cm}^2\ \checkmark\)
Common pitfalls
Frequently asked questions
What is the null factor law?
If the product of two factors is zero, then at least one of the factors is zero. So if \((x - 2)(x + 5) = 0\), then \(x = 2\) or \(x = -5\).
Why must one side of the equation be zero?
The null factor law only works for zero. If two factors multiply to \(8\), neither has to be any particular number, so you must rearrange to get zero first.
Why can't I divide both sides by x?
Because \(x\) might be zero, and dividing by zero is not allowed. Dividing by \(x\) throws away the solution \(x = 0\).
How many solutions can a quadratic equation have?
Two, one repeated solution, or none that are real numbers.
How do you solve an equation with x in the denominator?
Note that \(x\) cannot be zero, multiply every term by \(x\), rearrange to make one side zero, then factorise and solve.
What do I do with a negative answer in a word problem?
Check whether it makes sense. A negative length, mass or time is rejected; a negative number in a pure number problem may be valid.