# Quadratic Graphs – A Level Maths Revision

The good news is that, at A Level, there is no new content on Quadratic Graphs. However, you will need to have a thorough understanding of the GCSE content. But don’t worry, we’ll help you with that here… Continue reading Quadratic Graphs – A Level Maths Revision

# Top Ten Revision Techniques

Effective revision practice can save hours of time, relieve tiredness, increase motivation and enliven flat study routines. These top 10 revision techniques should help to prove these points as you seek to maximise your pre-exam preparation! Continue reading Top Ten Revision Techniques

# Solving Trigonometric Equations – A Level Maths Revision

There aren’t any complicated equations when it comes to sourcing your A Level Maths revision – simply sticking with Beyond will equal a great learning experience as proved with this post on solving trigonometric equations. Continue reading Solving Trigonometric Equations – A Level Maths Revision

# How to Factorise a Cubic – A Level Maths Revision

Whether you’re preparing for advanced maths study or in the midst of the mayhem, Beyond Maths can help you with tutorial-style posts like this one on how to factorise a cubic.

You’ll find step-by-step support on tried and trusted methodologies that are brought to life with worked examples, diagrams, grids, and practice questions that you can try out for yourself.

So let’s get into it…

To factorise a cubic expression, you need to do three things:

This all builds on what you learnt at GCSE. To practise some of this GCSE content, there’s a great resource here, along with multiple-choice quizzes to make sure you remember everything you need to use the factor theorem or polynomial division. Once you’re caught up with how to factorise cubics, try this challenge task.

### 1. Find a linear factor

A linear factor of a cubic will be in the form (x + b) or (ax + b). To factorise a cubic, you need the value of b (and sometimes a).

To find this factor, use the factor theorem. The factor theorem says that, for a function f(x):

• if $(x - b)$ is a factor, the $f(b) = 0$;
• if $(ax-b)$ is a factor then $f(\frac{b}{a})=0$.

For example, the cubic below has a factor of $(x + 2)$:

$f(x) = 2x^3 + 11x^2 + 17x + 6$

You can show this by calculating f(-2); if (x + 2) really is a factor, f(-2) will equal 0:

\begin{aligned} f(-2) &= 2(-2)^3 + 11(-2)^2 + 17(-2) + 6 \\ &= 2 \times -8 + 11 \times 4 + 17 \times -2 + 6 \\ &= -16 + 44 -34 + 6 \\ &=0 \end{aligned}

f(-2) = 0 so (x + 2) is a factor.

In that example, you were given a factor. What if you need to find one yourself?

Consider what happens when you expand a cubic:

$(x + 2)(x + 3)(x - 5) = (x^2 + 5x + 6)(x - 5) = x^3 - 19x - 30$

The constant term, -30, is the result of multiplying the constant term from each bracket:

$2 \times 3 \times -5 = -30$

Therefore, the constant in any linear factor must be a factor of the constant in the cubic expression. This gives you a set of starting values to test for potential factors.

Be careful – each factor could be positive or negative. Even if the constant term is positive, it could still be made up of negative factors (for example, -3 × -5 × 2 = 30).

Say you want to factorise the cubic f(x) = 2x3 – 3x2 – 23x + 12; start by considering the values of f(x) when x is each of the (positive and negative) factors of 12:

In this case, f(-3) = 0 and f(4) = 0.

According to the factor theorem, if (xb) is a factor then f(b) = 0.

Therefore, f(x) = 2x3 – 3x2 – 23x + 12 has factors of (x + 3) and (x – 4).

Normally, you would stop once you’ve found a single factor.

In some cases, you may be able to find all three factors this way and hence factorise the cubic directly, but this is not a reliable or efficient method of factorising; in expressions where the coefficient of x3 is not 0 it may be very time consuming and, where there is a repeated root (for example, (x + 2)2(x + 3)) you may spend time looking for a third factor where only 2 exist.

In some cubic expressions, all of the factors may be of the form (ax + b). In this case, you’ll have to apply the factor theorem to fractions, where the numerator is a factor of the constant term and the denominator is a factor of the coefficient of x3. For example, given a function where f($\frac{-2}{3}$) = 0, that function has a factor (3x + 2).

### 2. Divide the cubic by the factor to get a quadratic

Once you’ve found a linear factor, you need to divide the cubic expression by that factor. This will split the cubic into a linear expression (your factor) and a quadratic expression.

There are a number different ways to go about dividing a cubic expression by a linear one. There is no ‘best’ method – just use the one you are more comfortable with.

Consider the cubic from the example above, 2x3 – 3x2 – 23x + 12, and its factor, (x + 3).

We’ll use three different methods to divide this cubic expression by its linear factor – polynomial long division, the grid method and equating coefficients. You can choose which one you prefer.

#### a. Polynomial Long Division

Start as you would for normal short or long division:

Divide the first term of the cubic by the first term of your divisor (the linear expression):

Multiply your answer (2x2) by the divisor. Place this underneath the relevant powers of x:

Subtract this new line from the original cubic:

Continue as before – divide the first term of this new line by the first term of the divisor, then multiply your result by the divisor:

Subtract this new line from the line above:

Then repeat one more time:

At this point, the final two lines should be equal. If they are not equal, you’ve made a mistake – either in your long division or in finding your factor.

This gives you the result that:

$(2x^3 - 3x^2 - 23x + 12) \div (x + 3) = 2x^2 - 9x + 4$

Or, rearranged:

$2x^3 - 3x^2 - 23x + 12 = (x+3)(2x^2 - 9x + 4)$

#### b. The Grid Method

If you’re used to using the grid method for multiplying or for expanding brackets, you can use a grid and work backwards to divide. As before, we’ll divide 2x3 – 3x2 – 23x + 12 by x + 3.

First, draw a grid. To get a cubic, you multiply a linear expression by a quadratic one. You know the linear expression (the factor), but not the quadratic expression. For now, leave the coefficients blank:

Now, start to fill in the gaps. You have:

• two multiplications that could lead to an x2 term (the quadratic x2 term multiplied by 3 or the linear and quadratic x terms multiplied together);
• two multiplications that could lead to an x term (the quadratic x term multiplied by 3 and the quadratic constant multiplied by x);
• only one multiplication that could lead to an x3 term (the quadratic x2 term multiplied by x);
• only one multiplication that could lead to a constant term (the quadratic constant term multiplied by 3)

You therefore have enough information to add the constant and x3 terms into the grid:

You can see that the missing coefficient of x2 must be 2 and the missing constant must be 4:

This gives you have enough information to fill in two more gaps:

Finally, consider the green squares. The x2 term in the starting cubic expression was -3x2, so the terms in the two green squares must add to -3x2. This means the term in the empty green square must be -9x2.

Similarly, the x term in the starting cubic was -23x. Therefore, the term in the empty blue square must be -27x (as 4x – 27x = -23x).

You can use either of these two new values to find the value of the missing coefficient of x, -9, and hence find the quadratic expression that multiplies to 2x3 – 3x2 – 23x + 12:

$(x + 3)(2x^2 -9x + 4) = 2x^3-3x^2-23x+12$

#### c. Equating Coefficients

This final method is more algebraic. As before, we’ll divide 2x3 – 3x2 – 23x + 12 by its factor x + 3.

Like the grid method, this is based on the fact that a cubic can be made by multiplying a linear expression by a quadratic one. You have a linear expression, x + 3; you want to find the quadratic expression.

For now, write that quadratic in the form ax2 + bx + c. You can say that:

$2x^3 - 3x^2 - 23x + 12 = (x + 3)(ax^2 + bx + c)$

Multiplying out the brackets gives:

$2x^3 - 3x^2 -23x + 12 = ax^3 + 3ax^2 + bx^2 +3bx +cx +3x$

Now, compare the coefficients of x3 on the left side to the coefficients of x3 on the right, the coefficients of x2 on the left side to the coefficients of x2 on the right, and so on. This gives 4 simultaneous equations:

\begin{aligned}a &= 2 \\ 3a+b &= -3 \\3b+c &= -23\\3c &= 12 \end{aligned}

You now have the values of a (2) and c (4). To find the value of b, substitute the value of a into the second equation or the value of c into the third equation:

\begin{aligned} 3(2) + b &= -3 \\ b &= -9 \end{aligned}

You can now say:

$2x^3 - 3x^2 -23x + 12 = (x + 3)(2x^2 -9x + 4)$

You’re nearly finished – the cubic is now in the form of a linear expression multiplied by a quadratic expression. The final step is to factorise the quadratic.

Depending on the context of the question, bear in mind that the quadratic may not be factorisable – it may be worth checking the discriminant (here, the discriminant is 49, so the quadratic has two real roots).

At this stage, you should be confident factorising quadratics. Using whichever method you prefer will give you:

$2x^2 - 9x + 4 = (2x -1)(x-4)$

Now, you can write the cubic in fully factorised form:

$2x^3 - 3x2 -23x + 12 = (x+3)(2x-1)(x-4)$

### Can your cubic be factorised?

While all cubic expressions have a least one factor, if any factor(s) are irrational then the cubic cannot be fully factorised. Whether this is likely to be the case depends on the context of the question.

### Practice Questions

Here’s some to try yourself:

1) x3 – 4x2 – 7x + 10

2) 2x3 + 5x2 – 14x – 8

3) 3x3 + 14x2 + 3x – 36

4) x3 + 6x2 + 12x + 8

5) 6x3x2 – 4x – 1

6) 3x3 + 8x2 – 33x + 10

7) 21x – 4x3 – 4x2 – 9

8) 9x3 + 33x2 – 56x + 20

9) 6x3x2 – 46x – 15

10) 24x3 + 46x2 + 11x – 6

1) (x + 2)(x – 5)(x – 1)

2) (2x + 1)(x + 4)(x – 2)

3) (3x – 4)(x + 3)2

4) (x + 2)3

5) (3x + 1)(2x + 1)(x – 1)

6) (3x – 1)(x – 2)(x + 5)

7) (3 – 2x)(2x – 1)(x + 3)

8) (3x – 2)2(x + 5)

9) (3x + 1)(2x + 5)(x – 3)

10) (2x + 3)(4x – 1)(3x + 2)

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