3.03 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.

This form allows a few key features of the corresponding graph to be identified. First, the sign of the leading term $a$a determines whether the polynomial generally increases or decreases. Second, the $y$y-intercept of the graph is the coefficient $d$d. Also, with a quick calculation, the point of inflection can be located at the point where $x=-\frac{b}{3a}$x=−b3a​. 

Unfortunately, this form doesn't make it easy to determine its zeros, and this is why being able to find the factorised form for a given cubic 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.

In fact, a cubic equation can have $1$1, $2$2 or $3$3 solutions, and the number of solutions depends on the quadratic factor in the factorisation. Let $r$r be the root of a cubic. Then $\left(x-r\right)$(x−r) is a factor of the cubic and hence, the cubic function can be expressed in the form:

 $y=\left(x-r\right)\times\left(Ax^2+Bx+C\right)$y=(x−r)×(Ax2+Bx+C) 

Notice that there is one linear factor and one quadratic factor. This means that the polynomial could be further factorised depending on the factorisation of $Ax^2+Bx+C$Ax2+Bx+C. 

Some strategies for factorising cubics include:

• Looking for special patterns, such as the 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 a quadratic factor has been found, the techniques from previous chapters can be used to factorise this part of the equation, if possible.

Factorising special cases

Remember previously looking 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

The following examples demonstrate this process.

Practice questions

Question 1

Question 2

Factorising in pairs

Sometimes a cubic polynomial can be factorised using a pairing strategy (also called factorising by grouping).

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

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 one of the factors is known, say $\left(x-r\right)$(x−r), then use expansion to 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

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

Therefore, $x^3-6x^2+12x-35=\left(x-5\right)\left(Ax^2+Bx+C\right)$x3−6x2+12x−35=(x−5)(Ax2+Bx+C). 

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, after expanding on the right, this implies $A=1$A=1. 

Working through this example in full, and with some practice, you might recognise some short cuts.

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

Then solving for the coefficient of the $x^2$x2 term, evaluate $B$B:

Hence, the polynomial  $x^3-6x^2+12x-35$x3−6x2+12x−35 can be written 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 5

Factorisation by polynomial division

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

The following video demonstrates this process:

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, 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 (i.e. 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, 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, then the polynomial would have been successfully factorised into a linear factor and a quadratic factor.

Practice questions

Question 6

Question 7

Finding a linear factor

In the previous examples, the linear factor was always provided. What happens if one isn't given? 

Technology could be used to find a linear factor by inspecting the graph, or an important theorem called the remainder theorem.

The remainder theorem states that if a polynomial is divided by $\left(x-r\right)$(x−r), the remainder will be the polynomial evaluated at $x=r$x=r.

Remember that dividing a polynomial by a factor will give a remainder of zero. This provides 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 is zero, this means that $\left(x-a\right)$(x−a) is a factor

Good candidates for guessing 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 states that 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, try factors of $6$6 :$\left\{1,-1,2,-2,3,-3,6,-6\right\}${1,−1,2,−2,3,−3,6,−6}. Which would be a factor?

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

Practice questions

Question 8

Question 9