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Tutor: Bilal

## In a nutshell

A quadratic in the form $ax^2+bx+c$ can be factorised. You have already learnt how to factorise a quadratic when $a=1$ by finding two numbers to multiply to give $c$ and add to give $b$​. This lesson will also discuss how to factorise a quadratic expression where $a>1$.

## Factorise when $a=1$​​

A quadratic is in the form

$ax^2 + bx +c$​​

When $a=1$ we can factorise a quadratic as follows:

#### PROCEDURE

 ​1. Ensure the quadratic is in the form $ax^2 + bx + c$ and identify $a,b$ and $c$​​ 2. Set up the answer with two brackets and an $x$ at the start of each bracket, as follows $(x \qquad)(x \qquad)$​​ 3. Find two numbers which multiply to give $c$ and add to give $b$​​ 4. Fill these numbers in the double brackets

### Example 1

Factorise

$x^2 + 10x + 24$

The quadratic is in the correct form where $b=10$ and $c=24$.

$6 \times 4 = 24 \space and \space 6 + 4 = 10$, so the numbers are $6 \space and \space 4$.

$x^2 + 10x+24 = \underline{(x+6)(x+4)}$

## Factorise when $a>1$

​When $a>1$ you can factorise a quadratic as follows:

#### PROCEDURE

 ​1. Ensure the quadratic is in the form $ax^2+bx+c$, and identify $a, b$ and $c$. 2. Multiply $a$ and $c$ to give a new number $d$.​ 3. Think of two numbers, $m$ and $n$ which multiply to give $d$ and add to give $b$.​ ​4. Write the quadratic using $m$ and $n$ in the form $ax^2+mx+nx+c$. ​5. Factorise $ax^2+mx$ into single brackets. ​6. Factorise $nx+c$ into single brackets. ​7. Use both factorised single brackets expressions to find the double brackets expression. The first bracket consists of the terms outside the single brackets, and the second bracket consists of the expression inside the single brackets.

### Example 2

Factorise

$2x^2+13x+15$​​

The quadratic has $a=2, b=13$ and $c=15$. Multiply $a$ and $c$ to give $2 \times 15 = 30$.

Think of two numbers which will multiply to give $30$ and add to give $13$. The two numbers are $10$ and $3$, as $10 \times 3 = 30$ and $10+3 = 13$.

Write the quadratic expression using $10$ and $3$, and factorise.

\begin {aligned}&2x^2+13x+15 \\&\underbrace{2x^2 + 10x}_{\text{factorise}} + \underbrace{3x + 15}_{\text{factorise}} \\&2x(x+5) + 3(x+5)\end {aligned}​​

The two expressions inside both single brackets should match, in this case $x+5$.

The final answer consists of two brackets, the first bracket contains the terms on the outside of each bracket, $2x$ and $+3$. And the second bracket is the repeated expression inside the single brackets factorised.

$\underline{(2x+3)(x+5)}$​​

### Example 3

Factorise and solve

$3x^2+11x-4=0$​​

The quadratic has $a=3, b=11$ and $c=-4$. Multiply $a$ and $c$ to give $-12$.

Think of two numbers which will multiply to give $-12$ and add to give $+11$. The two numbers are $+12$ and $-1$, as $12\times -1 = -12$ and $12-1=11$.

Write the quadratic expression using $12$ and $-1$, and factorise.

\begin {aligned}3x^2+11x-4&=0 \\\underbrace{3x^2+12x}_{\text{factorise}}\space \underbrace{-1x-4}_{\text{factorise}}&=0 \\3x(x+4) -1(x+4)&=0 \\(3x-1)(x+4)&=0\end {aligned}​​

Now the quadratic is factorised, it can be solved. Find the solutions by making either bracket equal to $0$.

$\underline{x=\frac 1 3 \space or \space x = -4}$​​

## FAQs - Frequently Asked Questions

### What do double brackets mean?

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