Factorising Quadratics Worksheets | Questions and Revision | MME

# Factorising Quadratics Worksheets, Questions and Revision

Level 4-7

Quadratics are algebraic expressions that include the term, $x^2$, in the general form,

$ax^2 + bx + c$

Where $a, b$, and $c$ are all numbers. We’ve seen already seen factorising into single brackets, but this time we will be factorising quadratics into double brackets.

$(nx+m)(px+q)$

There are 2 main types of quadratics you will need to be able to factorise; one where $a=1$ and the other where $a\neq1$.

Make sure you are happy with the following topics before continuing.

Level 4-5

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## Take Note: The Factorising Trick

There is a quick trick to determine whether you should add a $+$ or $-$ sign to your brackets. There are three sub-types which we will go over here.

Sub-type (a): contains all positives

These quadratics contain all positive terms, e.g. $x^2 +5x + 6$. When factorised, both brackets will contain $\bf{\bf{\large{+}}}$.

$x^2 + 5x + 6 = (x+3)(x+2)$

Sub-type (b): $b$ is negative and $c$ is positive

These quadratics contain a negative $b$ value and a positive $c$ value. When factorised, both brackets will contain $\bf{\large{-}}$.

$x^2 -10x +21 = (x-7)(x-3)$

Sub-type (c): $c$ is negative.

If $c$ is negative, when factorised, one bracket will contain a $\bf{+}$ the other will contains a $\bf{\large{-}}$. The order or these will need to be determined. These are the hardest type and require the most thought.

$x^2 + 3x -18 = (x+6)(x-3)$

Level 4-5
Level 4-5

## Type 1: Factorising quadratics ($a=1$)

When we say $a=1$, we mean the number before $x^2$ in  $x^2+bx+c$ will be $1$ (typically we don’t write the $1$). Any number that appears before an $x$ term is called a coefficient, so in this case, $a$ is the coefficient of $x^2$ which has a value of $1$.

Example: Factorise the following quadratic into two brackets, $x^2\textcolor{blue}{-3}x+\textcolor{red}{2}$

Step 1: First we can write two brackets with an $x$ placed in each bracket.

$(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,) (x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)$

Step 2: We can identify that this is a sub-type (b) quadratic, meaning both brackets will contain $\large{-}$

$(x\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,) (x\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,)$

Step 3: We have to find two numbers which multiply to make $\textcolor{red}{2}$ and when added together make $\textcolor{blue}{-3}$.

We know both numbers will be negative.

$-2 \times -1 = \textcolor{red}{2}$

$-2 + -1 = \textcolor{blue}{-3}$

Finally add these numbers to the brackets.

$(x-2) (x-1)$

Level 4-5
Level 6-7

## Type 2: Factorising quadratics ($a> 1$)

In this instance the general form of the equation is $ax^2+bx+c$ where $a>1$

Example: Factorise the following quadratic $4x^2+\textcolor{blue}{3x}\textcolor{red}{-1}$

Step 1: When $a>1$ it makes things more complicated. It is not immediately obvious what the coefficient of each $x$ term should be. There are two possible options,

$(4x \kern{1 cm} ) (x \kern{1 cm} )$ or $(2x \kern{1 cm} ) (2x \kern{1 cm} )$

Step 2: We can identify that this quadratic is part of sub-type (c) meaning it can contain $+$ and $-$

This is most important for quadratic pairs which are non-symmetrically creating a third option, all three are shown below.

\begin{aligned}(4x \kern{0.4 cm} +\kern{0.4 cm} )&(x \kern{0.4 cm}-\kern{0.4 cm} )\\ (4x \kern{0.4 cm} -\kern{0.4 cm} )&(x \kern{0.4 cm}+\kern{0.4 cm} )\\(2x \kern{0.4 cm}+\kern{0.4 cm} )&(2x \kern{0.4 cm}-\kern{0.4 cm} )\end{aligned}

Step 3: We need to find two numbers which when multiplied make $\textcolor{red}{-1}$

$\textcolor{red}{-1}$ has only one factor.

$-1 \times 1 = -1$

Step 4: We need to find a combination which gives $\textcolor{blue}{3x}$

We can test our $3$ possibilities,

\begin{aligned}(4x+1)(x -1) &= 4x^2 -3x -1 \\(4x-1)(x+1) &= 4x^2 + 3x -1 \\(2x+1)(2x-1) &= 4x^2 -1\end{aligned}

As we can see $(4x-1) (x+1)$ gives the correct expansion and is therefore the answer.

Level 6-7

## Example 1: Factorising Simple Quadratics

Factorise $x^2 - x - 12$.

[2 marks]

Step 1: Draw empty brackets

$(x \kern{1 cm} ) (x \kern{1 cm} )$

Step 2: Identify sub-type (b)

$(x \kern{0.4cm} + \kern{0.4cm} ) (x \kern{0.4 cm} - \kern{0.4cm} )$

Step 3: We are looking for two numbers which multiply to make $\textcolor{red}{-12}$ and add to make $\textcolor{blue}{-1}$. Let’s consider some factor pairs of $-12$.

\begin{aligned}(-1)\times12&=-12 \,\,\text{ and } -1 + 12 = 11\\(-2)\times6&=-12 \,\,\text{ and } -2 + 6 = 4\\ (-6)\times2&=-12 \,\,\text{ and } -6 + 2 = -4\\ (-3)\times4&=-12 \,\,\text{ and } -3 +4 = 1\\ \textcolor{red}{(-4)\times3}&\textcolor{red}{=-12 }\,\,\text{ and } \textcolor{blue}{-4 + 3 = -1}\end{aligned}

We could keep going, but there’s no need because the last pair, $-4$ and $3$, add to make $-1$. This pair fills both criteria, (as highlighted above) so the factorisation of $x^2 - x - 12$ is

$(x - 4)(x + 3)$

Note: You can try expanding the double brackets to check your answer is correct. You should always get your original quadratic equation if you do this correctly.

Level 4-5

## Example 2: Factorising Harder Quadratics

Factorise $2x^2 + 7x + 3$.

[3 marks]

Step 1: Draw the empty brackets. Even though $a>1$ there is only one possible option this time.

$(2x \kern{1 cm} ) (x \kern{1 cm} )$

Step 2: Identify sub-type (a), meaning both brackets contain $+$.

$(2x \kern{0.4cm} + \kern{0.4cm} ) (x \kern{0.4 cm} + \kern{0.4cm} )$

Step 3: Find two numbers which multiply to give $\textcolor{red}{3}$

$3$ only has one factor.

$3 \times 1 = 3$

Step 4: Find the combination which gives $7x$

\begin{aligned}\textcolor{blue}{(2x \,\, +\,\, 1) (x\,\, + \,\,3) }&\textcolor{blue}{= 2x^2 + 7x +3} \\ (2x \,\, +\,\, 3) (x\,\, + \,\,1) &= 2x^2 +5x + 3\end{aligned}

As we can see, $(2x + 1)(x + 3)$, gives the correct expansion and is therefore the correct answer.

Level 6-7

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### Example Questions

We are looking for two numbers which add to make $1$ and multiply to make $-30$.

The factors of $30$ that satisfy theses two requirements are $5$ and $6$.

Therefore, the full factorisation of $a^2 + a - 30$ is

$(a - 5)(a + 6)$

We are looking for two numbers which add to make $-5$ and multiply to make $6$.

The factors of $6$ that satisfy theses two requirements are $-2$ and $-3$.

Therefore, the full factorisation of $k^2 - 5k + 6$ is

$(k - 2)(k - 3)$

We are looking for two numbers which add to make $7$ and multiply to make $12$.

The factors of $12$ that satisfy theses two requirements are $3$ and $4$.

Therefore, the full factorisation of $x^2 + 7x + 12$ is,

$(x + 3)(x + 4)$

In the quadratic, $a = 3$, $11$ is positive and $c$ is positive. We can set up the brackets as follows: $(3x \,\,\, + \,\,\,)(x \,\,\, + \,\,\,)$.

We are looking for two positive numbers which multiply to make $6$. The possible factors of $6$ are

\begin{aligned}(6)\times(1)&=6 \\ (3)\times(2)&=6 \end{aligned}

We now test all the combinations:

\begin{aligned}(3x+6)(x+1)&=3x^2+6x+3x+6 \\ (3x+1)(x+6)&= 3x^2+x+18x+6 \\ (3x+3)(x+2)&= 3x^2+6x+3x+6 \\ (3x+2)(x+3) &= 3x^2+9x+2x+6\end{aligned}

Hence the correct factorisation is $(3x+2)(x+3)$

We can see this is a sub-type (c) meaning it will contain both $+$ and $-$

Factors of $-6$

$(-1) \times 6 = -6$

$(-6) \times 1 = -6$

$(-2) \times 3 = -6$

$(-3) \times 2 = -6$

Lets find the options which give $- 5m$

$(2m+2)(2m-3) = 4m^2 - 2m - 6$

$(4m + 1)(m-6) = 4m^2 -23m-6$

$(4m-1)(m+6) = 4m^2 +23m -6$

$(4m+3)(m-2) = 4m^2 -5m -6$

We can see that last option with $+3$ and $-2$ is the correct combination.

This gives the final answer to be:

$(4m+3)(m-2)$

### Worksheets and Exam Questions

#### (NEW) Factorising Quadratics Exam Style Solutions - MME

Level 4-5 New Official MME

#### (NEW) Factorising Harder Quadratics Exam Style Questions - MME

Level 6-7 New Official MME

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