# Factorising Quadratics

## Factorising Quadratics Revision

**Factorising Quadratics**

**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.

**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)

**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)

**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.

**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.

**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.

## Factorising Quadratics Example Questions

**Question 1:** Factorise a^2 + a - 30

**[2 marks]**

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 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)

**Question 2:** Factorise k^2 - 5k + 6

**[2 marks]**

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)

**Question 3:** Factorise x^2 + 7x + 12

**[2 marks]**

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)

**Question 4:** Factorise 3x^2 + 11x + 6

**[4 marks]**

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)

**Question 5:** Factorise 4m^2 - 5m - 6

**[3 marks]**

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)