Solving Quadratics by Factorising
Solving a quadratic equation by factorising uses the null factor law: if a product is zero, one of the factors must be zero. So once \(3x^2+x-10=0\) is written as \((3x-5)(x+2)=0\), the solutions are \(x=\dfrac{5}{3}\) and \(x=-2\). In NSW Year 10 Advanced maths the equation often needs rearranging to equal zero first (NESA outcome MA5‑EQU‑P‑02).
You will solve monic and non-monic quadratics, rearrange equations that do not equal zero, and clear fractions, rejecting any value that makes a denominator zero. You will build a quadratic from its solutions, solve with a constant \(a\) as a coefficient, solve a number problem about reciprocals, and use a substitution when a bracket is repeated. It is part of Year 10 Advanced Quadratics.
Every question with a fully worked solution.
- Solving Quadratics by Factorising - Video - Solving Quadratic Equations by Factoring Watch
Theory
To solve a quadratic equation by factorising, make one side zero, factorise the other side, then set each factor equal to zero. This Year 10 Advanced maths page covers the null factor law, monic and non-monic quadratics, perfect squares, equations that must be rearranged first, equations with fractions and solutions that must be rejected.
A quadratic equation can be written as \(ax^2+bx+c=0\) with \(a\neq 0\). This is called standard form. It has at most two solutions, also called roots.
The null factor law: if \(A\times B=0\), then \(A=0\) or \(B=0\). It only works when one side is zero. If \(A\times B=10\), there are endless pairs that multiply to \(10\), so nothing can be concluded.
A quadratic is monic when \(a=1\), such as \(x^2-3x-10\). It is non-monic when \(a\neq 1\), such as \(3x^2+10x-8\).
A perfect square such as \((2x+5)^2=0\) gives only one solution, because both factors are the same. When the equation has a pronumeral in a denominator, any value that makes a denominator zero is excluded, even if the algebra produces it.
The null factor law and the factorising patterns you need:
| Type | Pattern | Example |
|---|---|---|
| Common factor | \(ax^2+bx = x(ax+b)\) | \(4x^2-6x = 2x(2x-3)\) |
| Monic trinomial | \(x^2+bx+c=(x+m)(x+n)\), \(mn=c\), \(m+n=b\) | \(x^2-x-20=(x-5)(x+4)\) |
| Non-monic (split the middle) | find \(m,n\) with \(mn=ac\), \(m+n=b\) | \(2x^2+7x+3=(2x+1)(x+3)\) |
| Difference of two squares | \(a^2-b^2=(a-b)(a+b)\) | \(9x^2-49=(3x-7)(3x+7)\) |
| Perfect square | \(a^2\pm 2ab+b^2=(a\pm b)^2\) | \(x^2-12x+36=(x-6)^2\) |
How to solve a quadratic by factorising
- Expand any brackets and collect every term on one side, so the other side is \(0\).
- Take out any common factor first, including a common number such as \(2\) or a common factor \(x\). Never divide it away.
- Factorise what is left: monic trinomial, split the middle term, difference of two squares or perfect square.
- Apply the null factor law: set each factor equal to \(0\) and solve the linear equations.
- Check each solution in the original equation and reject any value that makes a denominator zero.
Other question types
- Fractions: state the excluded values, multiply every term by the lowest common denominator, then solve the quadratic (Example 3). Cross-multiplying works when each side is a single fraction.
- A common bracket on both sides: in \((x+1)^2=3x(x+1)\), bring everything to one side and take out \((x+1)\). Dividing by \(x+1\) loses \(x=-1\).
- Letters as coefficients: treat the letter as a number and factorise as usual (Example 4).
- A parameter \(k\): factorise to find the solutions in terms of \(k\), then apply the condition, such as two solutions that differ by a given amount. Remember both orders, \(k-3\) and \(3-k\).
- Word problems: name the unknown, form the equation, solve, then reject any solution the situation does not allow, such as a negative length.
(i) \(3x^2+10x-8=0\)
(ii) \(4x^2+20x+25=0\)
(i) \(ac = 3\times(-8) = -24\). The two numbers with product \(-24\) and sum \(10\) are \(12\) and \(-2\). Split the middle term and group.
| \(3x^2+12x-2x-8\) | \(=\) | \(0\) |
| \(3x(x+4)-2(x+4)\) | \(=\) | \(0\) |
| \((3x-2)(x+4)\) | \(=\) | \(0\) |
| \(3x-2=0 \text{ or } x+4\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{2}{3}} \text{ or } \boldsymbol{x=-4}\) |
(ii) \(4x^2 = (2x)^2\), \(25 = 5^2\) and \(20x = 2\times 2x\times 5\), so this is a perfect square.
| \((2x+5)^2\) | \(=\) | \(0\) |
| \(2x+5\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{-\dfrac{5}{2}}\) |
Both factors are the same, so there is only one solution.
(i) \(x(3x-1)=14\)
(ii) \(5x^3=20x\)
(i) The right side is \(14\), not \(0\), so the null factor law cannot be used yet. Expand and move everything to one side.
| \(3x^2-x-14\) | \(=\) | \(0\) |
| \((3x-7)(x+2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{7}{3}} \text{ or } \boldsymbol{x=-2}\) |
Check \(ac=-42\): \(-7\times 6=-42\) and \(-7+6=-1\). That gives the split \(3x^2-7x+6x-14\).
(ii) Dividing both sides by \(x\) would lose the solution \(x=0\). Make one side zero and factorise.
| \(5x^3-20x\) | \(=\) | \(0\) |
| \(5x(x^2-4)\) | \(=\) | \(0\) |
| \(5x(x-2)(x+2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{0}, \ \boldsymbol{2} \text{ or } \boldsymbol{-2}\) |
(i) \(x-\dfrac{6}{x+1}=4\)
(ii) \(1+\dfrac{12}{x^2-4}=\dfrac{3}{x-2}\)
(i) \(x\neq -1\). Multiply every term by \(x+1\).
| \(x(x+1)-6\) | \(=\) | \(4(x+1)\) |
| \(x^2+x-6\) | \(=\) | \(4x+4\) |
| \(x^2-3x-10\) | \(=\) | \(0\) |
| \((x-5)(x+2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{5} \text{ or } \boldsymbol{x=-2}\) |
Neither value is \(-1\), so both are solutions.
(ii) \(x^2-4=(x-2)(x+2)\), so \(x\neq 2\) and \(x\neq -2\). Multiply every term by \((x-2)(x+2)\).
| \((x^2-4)+12\) | \(=\) | \(3(x+2)\) |
| \(x^2+8\) | \(=\) | \(3x+6\) |
| \(x^2-3x+2\) | \(=\) | \(0\) |
| \((x-1)(x-2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(1 \text{ or } x=2\) |
\(x=2\) makes two denominators zero, so it is rejected. The only solution is \(\boldsymbol{x=1}\).
(i) Find a quadratic equation with integer coefficients whose solutions are \(x=-\dfrac{3}{4}\) and \(x=2\).
(ii) Solve \(3x^2+5ax-2a^2=0\) for \(x\), where \(a\neq 0\).
(i) Turn each solution into a factor. \(x=-\dfrac{3}{4}\) gives \(4x=-3\), so \(4x+3=0\). \(x=2\) gives \(x-2=0\).
| \((4x+3)(x-2)\) | \(=\) | \(0\) |
| \(4x^2-8x+3x-6\) | \(=\) | \(0\) |
| \(\boldsymbol{4x^2-5x-6}\) | \(=\) | \(\boldsymbol{0}\) |
Any non-zero multiple, such as \(8x^2-10x-12=0\), has the same solutions.
(ii) Treat \(a\) as a number. Look for two terms with product \(3\times(-2a^2)=-6a^2\) and sum \(5a\): they are \(6a\) and \(-a\).
| \(3x^2+6ax-ax-2a^2\) | \(=\) | \(0\) |
| \(3x(x+2a)-a(x+2a)\) | \(=\) | \(0\) |
| \((3x-a)(x+2a)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{a}{3}} \text{ or } \boldsymbol{x=-2a}\) |
Common pitfalls
Frequently asked questions
What is the null factor law?
If two numbers multiply to give zero, at least one of them must be zero. So if \((x-3)(x+5)=0\), then \(x=3\) or \(x=-5\). It only works when one side of the equation is zero.
How do you solve a non-monic quadratic equation by factorising?
Make one side zero. Find two numbers whose product is \(ac\) and whose sum is \(b\). Split the middle term using them, factorise in pairs, then set each factor to zero.
Why does a perfect square give only one solution?
Both factors are identical, such as \((2x+5)(2x+5)\), so both give the same value of \(x\). On a graph the parabola just touches the \(x\)-axis instead of crossing it.
Why can't I divide both sides by x?
\(x\) might be zero, and dividing by zero is not allowed. Dividing loses the solution \(x=0\). Move every term to one side and take out \(x\) as a common factor instead.
Why do some solutions have to be rejected?
When you multiply both sides by an expression containing \(x\), the new equation can have a solution that makes that expression zero. That value makes a denominator of the original equation zero, so it is not a solution.
How do you find a quadratic equation from its solutions?
Turn each solution into a factor. \(x=\dfrac{4}{5}\) gives the factor \(5x-4\), which keeps the coefficients whole numbers. Multiply the factors and expand.