Factorising Quadratics Worksheets | Questions and Revision | MME

# Factorising Quadratics Worksheets, Questions and Revision

Level 4 Level 5 Level 6 Level 7

## What you need to know

Quadratics are algebraic expressions of the form

$ax^2 + bx + c$

Where $a, b$, and $c$ are all numbers. We’ve seen factorising a bunch of terms into one bracket by taking all the common factors outside the bracket – if you haven’t seen this, you should revise factorising. The following is an example of a quadratic with numbers.

$x^2 - x + 12$

Where there isn’t a number in front of the $x^2$ or $x$ like this example, then you know the number is a 1.

To factorise quadratics you need to find:

• two numbers which times together to make the constant term
• and add together to make the coefficient of $x$.

### Take Note:

Factorising quadratics is a useful technique to learn. Having a good understanding of quadratic equations is key and this topic is the first time you are likely to have come across a quadratic. The following topics which require this key knowledge include:

Factorise $x^2 - x - 12$.

We are looking for two numbers which multiply to make -12 and add to make -1. Let’s consider some factor pairs of -12.

$-12=(-1)\times 12,\text{ or }(-2)\times 6,\text{ or }(-3)\times 4,\text{ or }(-4)\times 3$

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 Quadratics – The Difference of Two Squares

Factorise $x^2 - 9$.

There’s no $x$ term in this quadratic. Therefore we’re looking for two numbers which multiply to make -9 and add to make 0.

For 2 numbers to add to make zero, they must be the same number but with different signs. Furthermore, two of the same numbers being multiplied to make -9 must correspond to a square root. The square root of 9 is 3, and indeed, the pair: 3 and -3 satisfy both criteria of adding to zero and multiplying to make -9. Therefore, the factorisation is

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

This special case is called the difference of two squares, because it always takes the form of $x^2$ takeaway some other square number.

Note: So, if you see a quadratic with no $x$ term, then you know that it can be factorised quickly – with the square root of the constant term in both brackets – one negative and one positive.

### Example 3: Factorising Harder Quadratics

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

To do this, we must firstly multiply the number in front of $x^2$ by the constant term, so in this case that is: $2 \times 3 = 6$. Then, we should look for two numbers which multiply to make 6, and add to make 7.

$6 = 2 \times 3, \text{ or } 1 \times 6$

1 and 6 satisfy both criteria. Now we must rewrite the quadratic with the $x$ split into two parts. In other words, because our pair is 1 and 6, we write the $7x$ term as $x + 6x$:

$2x^2 + x + 6x + 3$

Next, we have to treat the quadratic like two separate expressions and factorise the first two terms into their own single bracket, then factorise the last two terms into their own single bracket:

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

Note: If you’ve done this correctly, both brackets end up the same as shown above.

This means that we can now take the bracket itself, $(2x + 1)$, out as a factor to get

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

which is the full factorisation of the quadratic.

### 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,

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

Here we can take a factor of 2 out so that we have,

$2(x^2 - 6x +5)$

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

The factors of 5 that satisfy theses two requirements are -1 and -5.

Therefore, the full factorisation of $2x^2 -12x + 10$ is,

$2(x -1)(x -5)$

Firstly, $4 \times (-6) = -24$. So, we are looking for two numbers which multiply to make -24 and add to make -5. Let’s look at some factor pairs for -24.

$-24 = 1 \times (-24) \text{ or } 2 \times (-12)\text{ or } 3 \times (-8),$

We can stop here because $-8 + 3 = -5$ as desired, so -8 and 3 is our pair. So, now we can rewrite the quadratic, splitting up $-5m$ into $-8m + 3m$:

$4m^2 - 8m + 3m - 6$

Now, we factorise the first two terms into a single bracket and the last two terms into another single bracket to get

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

As per the plan, both of the brackets are the same. This means we can take out $(m - 2)$ as a factor of both terms in the expression. This leaves us with

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

Which is the full factorisation of the quadratic.

Level 4-5

Level 6-7

Level 4-5

Level 4-5

GCSE MATHS