# Factorising Quadratics When a ≠ 1

Exam-takers, note-makers and revision-breakers, there comes a time in GCSE Higher Level Maths when you need to demonstrate a competent understanding of factorising quadratics when $a$ ≠ 1.

If you’re sitting the higher paper for your GCSE, you need to be able to factorise quadratics in the form $ax^2+bx+c$. This is more difficult when $a$ is not 1.

In this blog, we will talk about two methods to factorise these expressions – the partition method and the grid method. We’ve included some practice questions too!

Check out our Expanding and Factorising bundle for exam-style questions on this topic an many more.

### The Partition Method

Let’s look at factorising $15x^2-x-6$.

The first step is to identify the values of $a$, $b$ and $c$. In this case, $a$ = 15, $b$ = -1 and $c$ = -6. Make sure you include any negative signs in these values.

Now, we find the value of $ac$: $ac$ = 15 × -6
= -90

We are looking for a pair of numbers that multiply together to give $ac$ (-90) and add together to give $b$ (-1). The easiest way to find the numbers is to write out the factors of -90:

The only pair that adds together to give -1 is 9 and -10.

We use these two numbers to split the middle term into two terms: 9 $x$ and -10 $x$. $15x^2-x-6=15x^2+9x-10x-6$

Next, we find the highest common factor of $15x^2$ and $9x$ and factorise it out of the first two terms. $3x(5x + 3) - 10x - 6$

Then, we do the same for the final two terms. It’s important the expression in the brackets is the same, $(5x + 3)$, so we take out a factor of -2. $3x(5x + 3) - 2(5x + 3)$

Since the expressions in the two brackets are the same, we can combine the terms in front of them to make our second bracket: $15x^2-x-6=(3x-2)(5x+3)$

This is our final answer! You can check it by expanding the brackets.

### The Grid Method

Let’s look at factorising the same quadratic: $15x^2-x-6$.

Just as with the partition method, the first step is to identify the values of $a$, $b$ and $c$. In this case, $a$= 15, $b$ = -1 and $c$ = -6. Make sure you include any negative signs in these values.

Now, we find the value of $ac$: $ac$ = 15 × -6
= -90

We are looking for a pair of numbers that multiply together to give $ac$ (-90) and add together to give $b$ (-1). The easiest way to find the numbers is to write out the factors of -90:

The only pair that adds together to give -1 is 9 and -10.

This is where the methods are different. We write the expression in a grid with the $x^2$ term in the top left corner and the constant term in the bottom right corner. We then use our two numbers, 9 and -10, to fill the remaining two boxes with 9 $x$ and -10 $x$.

We find the highest common factor of 15 $x^2$ and 9 $x$, which is 3 $x$. We write this to the left of the top row:

Then, we consider each term across the top and find the term we need to multiply 3 $x$ by to get this term. For example, 3 $x$ × 5 $x$ = 15 $x^2$, so we write 5 $x$ above 15 $x^2$.

Then, we do the same for the bottom row. We simply find what we multiply 5 $x$ by to get -10 $x$:

Our factorised expression is made up of 2 brackets – one from the terms across the top and the other from the terms to the side. $15x^2-x-6=(3x-2)(5x+3)$

Factorising quadratics when a ≠ 1 can take some time to master. Try out these tasks to reinforce your learning.

### Factorising Quadratics – Practice Questions

Have a go at these questions to practise your preferred method.

1. $6x^2+13x+5$

2. $21x^2+29x-10$

3. $36x^2-60x+25$

1. $(2x + 1)(3x + 5)$

2. $(7x - 2)(3x + 5)$

3. $(6x - 5)^2$ or $(6x - 5)(6x - 5)$

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