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Factoring Cubics with an identifiable factor


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. 

Why factorise a cubic function? 

Theoretically all cubic functions (with real coefficients) have at least one real root, say $r$r (sometimes quite difficult to find) and so every cubic function is expressible as $y=\left(x-r\right)\times Q\left(x\right)$y=(xr)×Q(x), where $Q\left(x\right)$Q(x) is a quadratic expression. 

This means that the number of roots the cubic function has will depend on the factorisation of $Q\left(x\right)$Q(x). This is why we learn to factorise these expressions.

Factorising in pairs

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

For example the polynomial $P\left(x\right)=4x^3-4x^2-9x+9$P(x)=4x34x29x+9 can be factorised as follows:

$P\left(x\right)$P(x) $=$= $4x^3-4x^2-9x+9$4x34x29x+9
  $=$= $4x^2\left(x-1\right)-9\left(x-1\right)$4x2(x1)9(x1)
  $=$= $\left(x-1\right)\left(4x^2-9\right)$(x1)(4x29)
  $=$= $\left(x-1\right)\left(2x-3\right)\left(2x+3\right)$(x1)(2x3)(2x+3)

This means that the function $y=4x^3-4x^2-9x+9$y=4x34x29x+9 can be rewritten as $y=\left(x-1\right)\left(2x-3\right)\left(2x+3\right)$y=(x1)(2x3)(2x+3). This function then has the three zeros $1,1\frac{1}{2},-1\frac{1}{2}$1,112,112

Factorisation when one linear factor is known

If we know that one of the factors is, say $\left(x-r\right)$(xr), then we can use synthetic 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.

Example 1:

Suppose we wish to factorise the polynomial $P\left(x\right)=4x^3-4x^2-9x+9$P(x)=4x34x29x+9 knowing that it has the linear factor $\left(x-1\right)$(x1). This is true because  $P\left(1\right)=0$P(1)=0, and thus, by the factor theorem, $\left(x-1\right)$(x1) is a factor.

So we use synthetic division to divide $4x^3-4x^2-9x+9$4x34x29x+9 by $\left(x-1\right)$(x1) as follows:

Coefficients of $P\left(x\right)$P(x) $4$4 $-4$4 $-9$9 $9$9
$1$1 $4$4 $0$0 $-9$9
Coefficients of Quotient $4$4 $0$0 $-9$9 $0$0

Therefore from the table, we have that $P\left(x\right)=\left(x-1\right)\left(4x^2-9\right)$P(x)=(x1)(4x29)

Now the quadratic factor is factorable into linear factors as a difference of two squares, so that the complete factorisation becomes $P\left(x\right)=\left(x-1\right)\left(2x-3\right)\left(2x+3\right)$P(x)=(x1)(2x3)(2x+3)

Example 2:

The function $y=\left(x-2\right)^3-27$y=(x2)327 can be thought of as the base function $y=x^3$y=x3 translated so that its horizontal inflection sits on the point $\left(2,-27\right)$(2,27)

Alternatively the function can be expressed as a factorised cubic. By making use of the difference of two cubes identity $\left(a-b\right)\left(a^2+ab+b^2\right)$(ab)(a2+ab+b2), we have:

$y$y $=$= $\left(x-2\right)^3-27$(x2)327
  $=$= $\left[\left(x-2\right)-3\right]\left[\left(x-2\right)^2+3\left(x-2\right)+9\right]$[(x2)3][(x2)2+3(x2)+9]
  $=$= $\left(x-5\right)\left(x^2-4x+4+3x-6+9\right)$(x5)(x24x+4+3x6+9)
  $=$= $\left(x-5\right)\left(x^2-x+7\right)$(x5)(x2x+7)

The quadratic factor is irreducible (the discriminant $\Delta=b^2-4ac=-27$Δ=b24ac=27) and this means that the function has only one real zero at $x=5$x=5

Example 3:

The cubic polynomial $P\left(x\right)=6x^3-29x^2-7x+10$P(x)=6x329x27x+10 is known to have the factor $\left(2x-1\right)$(2x1). We can use synthetic division to find the other factors as follows:

Coefficients of $P\left(x\right)$P(x) $6$6 $-29$29 $-7$7 $10$10
$\frac{1}{2}$12   $3$3 $-13$13 $-10$10
$2\times$2× Coefficients of Quotient $6$6 $-26$26 $-20$20 $0$0


Thus $P\left(x\right)=\left(2x-1\right)\left(3x^2-13x-10\right)$P(x)=(2x1)(3x213x10) and the discriminant of the quadratic can be determined as $\Delta=b^2-4ac=49$Δ=b24ac=49 . This means that the quadratic factor will break up into two linear factors. 

The fully factorised polynomial becomes $P\left(x\right)=\left(2x-1\right)\left(3x+2\right)\left(x-5\right)$P(x)=(2x1)(3x+2)(x5)

The graph of the function given by $y=\left(2x-1\right)\left(3x+2\right)\left(x-5\right)$y=(2x1)(3x+2)(x5) has $x$x-intercepts of $\frac{1}{2},-\frac{2}{3}$12,23 and $5$5.

More Worked Examples


The polynomial $x^3+5x^2+2x-8$x3+5x2+2x8 has a factor of $x-1$x1.

  1. By observation, find the quadratic factor of $x^3+5x^2+2x-8$x3+5x2+2x8.


  2. Hence factorise the polynomial completely.


The polynomial $4x^3+21x^2+29x+6$4x3+21x2+29x+6 has a factor of $x+3$x+3.

  1. By observation, find the quadratic factor of $4x^3+21x^2+29x+6$4x3+21x2+29x+6.


  2. Hence factorise the polynomial completely.


Consider the expression $x^3-64$x364.

  1. Factorise this expression.

  2. Does $x^3-64$x364 have any other real-valued linear factors?




  3. State the number of real-valued linear factors that $x^3-64$x364 has.

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