# 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

## 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 Level

## 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 Level

## 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 Level
A Level

## 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 Level

## 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:

A Level

A Level

A Level

A Level

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