1.09 Factorise cubic polynomials

The 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 though is immediately determine its zeros, 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. Say $r$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 remainder 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. Let's look at these now.

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

Now let's look at some examples and 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

Suppose we wish to factorise the polynomial $y=x^3-6x^2+12x-35$y=x3−6x2+12x−35 and have been given it has the linear factor $\left(x-5\right)$(x−5). 

Therefore, 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. 

Let's do this example in full and with some practice you might recognise some short cuts.

$x^3-6^2+12x-35$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)

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 4

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

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

The dividend: $3x^3+2x^2-6$3x3+2x2−6

The divisor: $x+2$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.

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

Practice questions

Question 5

Question 6

 

 

The remainder theorem

To divide two polynomials, say $P\left(x\right)$P(x) divided by $D\left(x\right)$D(x), we need the degree of the divisor polynomial $D\left(x\right)$D(x) to be less than or equal to the degree of the dividend polynomial $P\left(x\right)$P(x). 

As an example, suppose we divide $P\left(x\right)=x^2-5x+6$P(x)=x2−5x+6 by the polynomial $D\left(x\right)=x-1$D(x)=x−1. We proceed in a manner similar to long division given as follows:

Thus we state:

$\frac{P\left(x\right)}{D\left(x\right)}=\frac{x^2-5x+6}{x-1}=\left(x-4\right)+\frac{2}{x-1}$P(x)D(x)​=x2−5x+6x−1​=(x−4)+2x−1​

We can express the result slightly differently by multiplying both sides by the divisor so that:

$P\left(x\right)=x^2-5x+6=\left(x-1\right)\left(x-4\right)+2$P(x)=x2−5x+6=(x−1)(x−4)+2

Stating the result like this is known as the division transformation.

In general, dividing $P\left(x\right)$P(x) by $\left(x-a\right)$(x−a) will always produce a result that looks like:

$P\left(x\right)=\left(x-a\right)Q\left(x\right)+R$P(x)=(x−a)Q(x)+R

Here, $Q\left(x\right)$Q(x) is the quotient polynomial of one less degree than $P\left(x\right)$P(x) and $R$R is the remainder.

This last equation holds the key to the remainder theorem. Because the polynomial holds true for all values of $x$x, by putting $x=a$x=a into this general result we see that:

$P\left(a\right)=\left(a-a\right)Q\left(a\right)+R=R$P(a)=(a−a)Q(a)+R=R

This means that substituting $x=a$x=a into $P\left(x\right)$P(x) before dividing will reveal the remainder. It's a little mathematical magic! We can actually know the remainder even before the division by $\left(x-a\right)$(x−a) is done. This nice result is known as the remainder theorem.

Let's try it with our example. With $P\left(x\right)=x^2-5x+6$P(x)=x2−5x+6 and $D\left(x\right)=x-1$D(x)=x−1, before dividing note that $P\left(1\right)=\left(1\right)^2-5\left(1\right)+6=2$P(1)=(1)2−5(1)+6=2 and this is indeed the remainder!

 

The factor theorem

The factor theorem is an extension of the remainder theorem. 

If a polynomial equation $P(x)=0$P(x)=0 has a root $x=a$x=a, meaning $P(a)=0$P(a)=0, then $x-a$x−a must be a factor of $P(x)$P(x). We could write $P(x)=(x-a)Q(x)$P(x)=(x−a)Q(x) where $Q$Q is a polynomial of degree one less than the degree of $P$P.

This means that if we can find by any means a number $a$a such that $P(a)=0$P(a)=0, then we know immediately that $x-a$x−a is a factor of $P$P.

 

Finding linear factors

We are often asked to find a linear factor of $p\left(x\right)$p(x). Using trial and error, we will substitute in values until we find a factor, that is a value such that $p\left(a\right)=0$p(a)=0.

For $p\left(x\right)=ax^n+...+c$p(x)=axn+...+c, we should start our trial and error with factors of $c$c and then move on to $\frac{\text{factors of }c}{\text{factors of }a}$factors of cfactors of a​.

For example, for the polynomial $p\left(x\right)=x^3-x^2-x-2$p(x)=x3−x2−x−2 we would want to try $x=1$x=1, $x=-1$x=−1, $x=2$x=2 and$x=-2$x=−2.

 

Worked example

example 2

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, $p(x)=(x-2)q(x)$p(x)=(x−2)q(x).

Do: We can use the division algorithm to calculate $q(x)$q(x). That is, we divide $p(x)$p(x) by $x-2$x−2. In this way, we find that $q(x)=x^2+x+1$q(x)=x2+x+1 which is a prime polynomial. So, $p(x)=x^3-x^2-x-2=(x-2)(x^2+x+1)$p(x)=x3−x2−x−2=(x−2)(x2+x+1) is the fully factored form of $p\left(x\right)$p(x).

 

Practice questions

Question 7

 

Question 8

Question 9