Cubics

Cubics

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

Cubics

A cubic is an expression in the form \textcolor{red}{a}x^3 + \textcolor{blue}{b}x^2 + \textcolor{limegreen}{c}x + \textcolor{orange}{d}. You can factorise cubics, in a similar way to factorising quadratics – however it it a little trickier. There are 2 methods used to factorise cubics.

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

A Level AQA Edexcel OCR

Method 1: Factorising Cubics given 1 Factor

When factorising a cubic expression, you will be able to put it into (up to) 3 brackets. You may be given 1 factor that you can work with, which will make the factorisation easier.

Example: (x-2) is a factor of f(x) = 2x^3 + x^2 - 13x + 6. Hence, express f(x) as a product of three linear factors.

Step 1: Write down the factor we know, \textcolor{orange}{(x-2)}, next to another set of brackets:

\textcolor{orange}{(x-2)}( \quad \quad \quad \quad \quad \quad ) = 2x^3 + x^2 - 13x + 6

Step 2: Put an x^2 term at the start of the second set of brackets – this will need to multiply together with x from the first bracket to make 2x^3:

\textcolor{orange}{(x-2)}(\textcolor{red}{2x^2} \quad \quad \quad \quad ) = 2x^3 + x^2 - 13x + 6

Step 3: Find the constant term for the end of the second bracket – this will need to be multiplied together with -2 to make 6:

\textcolor{orange}{(x-2)}(\textcolor{red}{2x^2} \quad \quad \quad \textcolor{limegreen}{-3}) = 2x^3 + x^2 - 13x + 6

Step 4: The x and -3 above multiply to make -3x, but we need -13x. So we need an extra -10x from somewhere – we need to multiply -2 by 5x to make -10x:

\textcolor{orange}{(x-2)}(\textcolor{red}{2x^2} + \textcolor{blue}{5x} \textcolor{limegreen}{-3}) = 2x^3 + x^2 - 13x + 6

Step 5: Check that this works, by multiplying out the brackets and seeing if this matches f(x):

\begin{aligned} &\, \textcolor{orange}{(x-2)}(\textcolor{red}{2x^2} + \textcolor{blue}{5x} \textcolor{limegreen}{-3}) \\ &= 2x^3 + 5x^2 - 3x - 4x^2 - 10x + 6 \\ &= 2x^3 + x^2 - 13x + 6 \end{aligned}

So, this works so far.

Step 6: Factorise the quadratic 2x^2 + 5x -3 into 2 linear factors:

\textcolor{red}{2x^2} + \textcolor{blue}{5x} \textcolor{limegreen}{-3} = \textcolor{orange}{(2x-1)(x+3)}

Step 7: Put it all together:

2x^3 + x^2 - 13x + 6 = \textcolor{orange}{(x-2)(2x-1)(x+3)}

A LevelAQAEdexcelOCR

Method 2: Factorising Cubics using the Factor Theorem

If you are given no factors, then you can use the Factor Theorem to find one factor, and then use Method 1 from above.

Reminder: The Factor Theorem is defined as:

“If f(x) is a polynomial, and f(k)=0, then (x-k) is a factor of f(x)

or

“If f \left(\dfrac{b}{a} \right)=0, then (ax-b) is a factor of f(x)

 

When using the factor theorem here, you will need to try small values of k, e.g. f(1), f(-1), f(2), etc., until you find f(k)=0.

A LevelAQAEdexcelOCR

Cubic Graphs

You can sketch a cubic if you know its factors. You have to find where the function is 0.

All cubic graphs have a general shape:

  • If the coefficient of x^3 is positive, then the graph goes from ‘the bottom left to the top right’
  • If the coefficient of x^3 is negative, then the graph goes from ‘the top left to the bottom right’

The y-intercept of a cubic graph y=ax^{3}+bx^{2}+cx+d is always d, so by expanding brackets we can find where the graph crosses the y-axis.

 

Example: Sketch the graphs of:

\begin{aligned} f(x) &= x(x-1)(x+2) \\ g(x) &= (x+1)(x^2 + x + 1) \\ h(x) &= (1-x)(x+2)^2 \\ m(x) &= (1-2x)^3 \end{aligned}

A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example: Factorising Cubics using the Factor Theorem

Factorise f(x) = 3x^3 - 5x^2 - 4x + 4

[4 marks]

Try small values of k, e.g. f(1), f(-1), f(2), etc., until we find f(k)=0:

\begin{aligned} f(1) &= 3(1)^3 - 5(1)^2 - 4(1) + 4 = -2 \\ f(-1) &= 3(-1)^3 - 5(-1)^2 - 4(-1) + 4 = 0 \end{aligned}

f(-1) = 0, hence \textcolor{orange}{(x+1)} is a factor of f(x), by the Factor Theorem.

 

Write down the factor we know, \textcolor{orange}{(x+1)}, next to another set of brackets:

\textcolor{orange}{(x+1)}( \quad \quad \quad \quad \quad \quad ) = 3x^3 - 5x^2 - 4x + 4

 

Then, using Method 1, factorise the cubic:

We need the x^{2} term in our brackets to multiply by x to get 3x^{3}, so it must be 3x^{2}.

\textcolor{orange}{(x+1)}( 3x^{2} \quad \quad \quad \quad \quad ) = 3x^3 - 5x^2 - 4x + 4

1\times3x^{2}=3x^{2} but we want -5x^{2}, so we need -8x to multiply the x in the orange bracket to give us the difference of -8x^{2}.

\textcolor{orange}{(x+1)}( 3x^{2}-8x \quad \quad \quad ) = 3x^3 - 5x^2 - 4x + 4

To get the 4 at the end we multiply the 1 in the first bracket by the number that will go at the end of the second bracket, so this must be 4.

\textcolor{orange}{(x+1)}( 3x^{2}-8x+4) = 3x^3 - 5x^2 - 4x + 4

Finally, factorise the second bracket.

\textcolor{orange}{(x+1)}( 3x-2)(x-2) = 3x^3 - 5x^2 - 4x + 4

A LevelAQAEdexcelOCR

Example Questions

Write down the factor we know, (2x+1), next to another set of brackets:

 

(2x+1)( \quad \quad \quad \quad \quad \quad ) = 2x^3 + 5x^2 - 4x - 3

 

Put an x^2 term at the start of the second set of brackets – this will need to multiply together with 2x from the first bracket to make 2x^3:

 

(2x+1)(x^2 \quad \quad \quad \quad ) = 2x^3 + 5x^2 - 4x - 3

 

Find the constant term for the end of the second bracket – this will need to be multiplied together with 1 to make -3:

 

(2x+1)(x^2 \quad \quad \quad -3) = 2x^3 + 5x^2 - 4x - 3

 

The 2x and -3 above multiply to make -6x, but we need -4x. So we need an extra +2x from somewhere – we need to multiply 1 by 2x to make +2x:

 

(2x+1)(x^2 + 2x - 3) = 2x^3 + 5x^2 - 4x - 3

 

Check that this works, by multiplying out the brackets and seeing if this matches f(x):

 

\begin{aligned} &\, (2x+1)(x^2 + 2x - 3) \\ &= 2x^3 + 4x^2 - 6x + x^2 + 2x - 3 \\ &= 2x^3 + 5x^2 - 4x - 3 \end{aligned}

 

So, this works so far.

Factorise the quadratic x^2 + 2x -3 into 2 linear factors:

 

x^2 + 2x - 3 = (x+3)(x-1)

 

Finally, put it all together:

2x^3 + 5x^2 - 4x - 3 = (2x+1)(x+3)(x-1)

Try small values of k until f(k) = 0:

 

\begin{aligned} f(1) &= 3(1)^3 - (1)^3 - 20(1) - 12 = -30 \\ f(-1) &= 3(-1)^3 - (-1)^3 - 20(-1) - 12 = 4 \\ f(2) &= 3(2)^3 - (2)^2 - 20(2) - 12 = -32 \\ f(-2) &= 3(-2)^3 - (-2)^2 - 20(-2) - 12 = 0 \end{aligned}

 

f(-2) = 0, so (x+2) is a factor, by the Factor Theorem.

 

Then, write down this set of brackets next to a set of empty brackets, and factorise using the usual method:

 

\begin{aligned} (x+2)( \quad \quad \quad \quad \quad \quad ) &= 3x^3 - x^2 - 20x - 12 \\ 3x^3 - x^2 - 20x - 12 &= (x+2)(3x^2 - 7x - 6) \\ &= (x+2)(3x+2)(x-3) \end{aligned}

a) (3x-1) is a factor of f(x) if f \left( \dfrac{1}{3} \right) = 0, by the Factor Theorem:

 

f \left( \dfrac{1}{3} \right) = 3 \left( \dfrac{1}{3} \right)^3 - 10 \left(\dfrac{1}{3} \right)^2 + 9 \left( \dfrac{1}{3} \right) - 2 = 0

 

Hence, (3x-1) is a factor of f(x)

 

b)

\begin{aligned} 3x^3 - 10x^2 + 9x - 2 &= (3x-1)(x^2 - 3x + 2) \\ &= (3x-1)(x-1)(x-2) \end{aligned}

 

c) When f(x) = (3x-1)(x-1)(x-2) = 0, x = \dfrac{1}{3}, x = 1 or x = 2

Hence, the graph will cross the x-axis at x = \dfrac{1}{3}, x = 1 and x = 2

 

f(0) = 3(0)^3 - 10(0)^2 + 9(0) - 2 = -2

Hence, the graph will cross the y-axis at y = -2

 

So, we have enough information to sketch the graph:

 

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