1.09 Factorise cubic polynomials

A cubic function can be written down in standard polynomial form as $y=ax^3+bx^2+cx+d$y=ax3+bx2+cx+d.

In this form we can immediately notice a few things about the function's graph. Notwithstanding the presence of local maxima and minima, we know from the sign of the leading term $a$a whether it generally increases or decreases. We also know its $y$y-intercept (the coefficient $d$d). Also with a quick calculation, we can locate the point of inflection at the point where $x=-\frac{b}{3a}$x=−b3a​. 

What we can't do immediately is determine its zeros ($x$x-intercepts), and this is why the function's factorised form is so important.

 

How to factorise cubics 

Theoretically all cubic functions (with real coefficients) have at least one real root. Visualise the graph of a cubic equation, it must have at least one $x$x-intercept and therefore, at least one real root. A cubic equation can have $1$1, $2$2 or $3$3 solutions, the number depends on the quadratic in the factorisation. If $r$r is a root of a cubic then we know $\left(x-r\right)$(x−r) is a factor of the cubic and hence, we could express the cubic function in the form:

 $y=\left(x-r\right)\times\left(Ax^2+Bx+C\right)$y=(x−r)×(Ax2+Bx+C), we can see we have one linear factor and a quadratic factor. 

This means that we could then further factorise the cubic depending on the factorisation of $Ax^2+Bx+C$Ax2+Bx+C. 

Some strategies for factorising cubics include:

• Looking for special patterns, such as difference of two cubics, sum of two cubics
• Factorise by grouping in pairs
• Given one linear factor, find the quadratic factor by expansion and equating coefficients
• Given one linear factor, find the quadratic factor by polynomial division
• Find a linear factor using the factor theorem or technology and then find the quadratic factor using one of the previous two methods above

Once you have found a quadratic factor you can use our techniques from previous chapters to factorise this part of the equation if possible.

Let's look at examples of each of these techniques.

 

Factorising special cases

Previously we've looked at some special factorising rules, such as the difference of two squares. It is useful to be able to spot a few special cases for cubics too.

Some people use the mnemonic "SOAP" to help remember the order of the signs in these formulae. The letters stand for:

SAME as the sign in the middle of the original expression

OPPOSITE sign to the original expression

ALWAYS POSITIVE

Work through these practice questions to see this process in action!

 

Practice questions

Question 1

Question 2

 

Factorising in pairs

Sometimes a cubic polynomial can be factorised using a pairing strategy.

For example the polynomial $y=4x^3-4x^2-9x+9$y=4x3−4x2−9x+9 can be factorised as follows:

$y$y	$=$=	$4x^3-4x^2-9x+9$4x3−4x2−9x+9
	$=$=	$4x^2\left(x-1\right)-9\left(x-1\right)$4x2(x−1)−9(x−1)
	$=$=	$\left(x-1\right)\left(4x^2-9\right)$(x−1)(4x2−9)
	$=$=	$\left(x-1\right)\left(2x-3\right)\left(2x+3\right)$(x−1)(2x−3)(2x+3)

This means that the function $y=4x^3-4x^2-9x+9$y=4x3−4x2−9x+9 can be rewritten as $y=\left(x-1\right)\left(2x-3\right)\left(2x+3\right)$y=(x−1)(2x−3)(2x+3). This function then has the three zeros $1,\frac{3}{2},-\frac{3}{2}$1,32​,−32​. 

 

Practice question

Question 3

 

Factorisation when one linear factor is known

If we know that one of the factors is, say $\left(x-r\right)$(x−r), then we can use expansion and equate coefficients or use polynomial division to find the other factor or factors. Remember that a cubic polynomial will factorise into a linear factor and a quadratic factor. Then, the quadratic factor may (or may not) be able to be factorised into two linear factors itself.

 

Worked example

Example 1

Factorise the polynomial $y=x^3-6x^2+12x-35$y=x3−6x2+12x−35 , given it has the linear factor $\left(x-5\right)$(x−5). 

Think: We have that $x^3-6x^2+12x-35=\left(x-5\right)\left(Ax^2+Bx+C\right)$x3−6x2+12x−35=(x−5)(Ax2+Bx+C). 

We can expand the right hand side and equate coefficients. This can often be done quickly by observation. For example, on the right the coefficient of $x^3$x3 is $1$1 and if we expand on the right we can see that this means $A=1$A=1. 

Do:

$x^3-6^2+12x-35$x3−62+12x−35	$=$=	$\left(x-5\right)\left(Ax^2+Bx+C\right)$(x−5)(Ax2+Bx+C)
	$=$=	$Ax^3+Bx^2+Cx-5Ax^2-5Bx-5C$Ax3+Bx2+Cx−5Ax2−5Bx−5C
	$=$=	$Ax^3+\left(B-5A\right)x^2+\left(C-5B\right)x-5C$Ax3+(B−5A)x2+(C−5B)x−5C	By collecting like terms.

Equating coefficients on the left with those on the right we can first solve for $A$A and $C$C:

$A$A	$=$=	$1$1	and	$-5C$−5C	$=$=	$-35$−35
				$C$C	$=$=	$7$7

Then solving for the coefficient of the $x^2$x2 term we can find $B$B:

$B-5$B−5	$=$=	$-6$−6
$B$B	$=$=	$-1$−1

Hence, we can write  $x^3-6x^2+12x-35$x3−6x2+12x−35 in the form $\left(x-5\right)\left(x^2-x+7\right)$(x−5)(x2−x+7)

Reflect: The quadratic factor here cannot be further factorised into linear factors with real roots (the discriminant $b^2-4ac=-27$b2−4ac=−27) and this means that the function has only one real zero at $x=5$x=5.

 

Practice question

Question 5

Factorisation by polynomial division

We can divide one polynomial by another polynomial of the same or lower degree, using a process similar to long division. If we know a linear factor $\left(x-r\right)$(x−r) of a cubic $ax^3+bx^2+cx+d$ax3+bx2+cx+d, we can use polynomial division to find the quadratic factor. 

Let's look at the process, then try some examples.

 

The process using long division

Worked example

example 2

Solve: $\left(3x^3+2x^2-6\right)\div\left(x+2\right)$(3x3+2x2−6)÷​(x+2) using long division

Think: The dividend is $3x^3+2x^2-6$3x3+2x2−6 and the divisor is $x+2$x+2

It's helpful to write the dividend like this: $3x^3+2x^2+0x-6$3x3+2x2+0x−6. Then we can use long division to solve the problem.

Do:

1. Divide the first term of the dividend by the highest term of the divisor (meaning the one with the highest power of $x$x, which in this case is $x$x). Place the result above the bar ($3x^3\div x=3x^2$3x3÷​x=3x2). 

2. Multiply the divisor by the number you just wrote above the top line. Write the result under the first two terms of the dividend. ($3x^2\times\left(x+2\right)=3x^3+6x^2$3x2×(x+2)=3x3+6x2). 

3. Subtract these terms from the from those in the original dividend, making sure you pay attention to the positive and negative signs ($3x^3+2x^2-\left(3x^3+6x^2\right)=-4x^2+0x$3x3+2x2−(3x3+6x2)=−4x2+0x). 

4. Repeat the previous three steps, except this time use the two terms that have just been written as the dividend (ie. divide $-4x^2+0x$−4x2+0x by $x+2$x+2). 

5. Repeat step 4. Keep going until there is nothing to "pull down".

The polynomial above the bar is the quotient, and the number left over, $-22$−22, is the remainder.

From this we can write: $3x^3+2x^2-6=\left(x+2\right)\left(3x^2-4x+8\right)-22$3x3+2x2−6=(x+2)(3x2−4x+8)−22.

Reflect: If the remainder was zero we would have successfully factorised the cubic expression into a linear factor and a quadratic factor.

 

Practice questions

Question 6

Question 7

 

 

The factor theorem

Consider the cubic function $P(x)=x^3+x^2-4x+1$P(x)=x3+x2−4x+1.

In factorised form this can be written as $P(x)=(x-2)(x+2)(x+1)$P(x)=(x−2)(x+2)(x+1)

The roots of the function are $2$2, $-2$−2 and $-1$−1. $P(2)=0$P(2)=0 and $(x-2)$(x−2) is a factor of $P(x)$P(x),  $P(-2)=0$P(−2)=0 and $x+2$x+2 is a factor of $P(x)$P(x), $P(-1)=0$P(−1)=0 and $x+1$x+1 is a factor of $P(x)$P(x).

If $r$r is a root of a polynomial then we can say that $x-r$x−r must be a factor. This is called the factor theorem which is part of the remainder theorem for polynomials. 

 

Finding a linear factor

So far we have been given a linear factor but how could we find one if it wasn't given? 

The factor theorem gives us a great way to find factors using a guess and check method:

• Guess a root of the expression, say $a$a
• Try evaluating the expression for this value
• If the result zero, you have found a factor $\left(x-a\right)$(x−a)

Good candidates for guesses for a root of the expression $ax^3+bx^2+cx+d$ax3+bx2+cx+d would be factors of $d$d divided by factors of $a$a. This comes from the rational root theorem, which tells us if there are rational roots they will be of the form $\frac{p}{q}$pq​, where $p$p is a factor of $d$d and $q$q is a factor of $a$a. For example, when trying to find roots for the cubic $x^3+3x-13x+6$x3+3x−13x+6, we could try factors of $6$6 :$\left\{1,-1,2,-2,3,-3,6,-6\right\}${1,−1,2,−2,3,−3,6,−6}. Can you find a factor?

It may not always be possible to find some linear factors quickly using this method. 

Worked example

example 3

Factorise $p(x)=x^3-x^2-x-2$p(x)=x3−x2−x−2.

Think: Using the factor theorem, we know we are looking for a value $a$a such that $p\left(a\right)=0$p(a)=0. By trial and error, we find that $p(2)=0$p(2)=0. Therefore, $x-2$x−2 is a factor and $p(x)=(x-2)q(x)$p(x)=(x−2)q(x).

Do: Use the division algorithm to calculate $q(x)$q(x). That is, we divide $p(x)$p(x) by $x-2$x−2.

We find that $q(x)=x^2+x+1$q(x)=x2+x+1 which cannot be factorised further. So we write$p(x)=x^3-x^2-x-2=(x-2)(x^2+x+1)$p(x)=x3−x2−x−2=(x−2)(x2+x+1)

 

Practice questions

Question 8

Question 9