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=(x−r)×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)=4x3−4x2−9x+9 can be factorised as follows:
| $P\left(x\right)$P(x) | $=$= | $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 $1,1\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 $\left(x-r\right)$(x−r), 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)=4x3−4x2−9x+9 knowing that it has the linear factor $\left(x-1\right)$(x−1). This is true because $P\left(1\right)=0$P(1)=0, and thus, by the factor theorem, $\left(x-1\right)$(x−1) is a factor.
So we use synthetic division to divide $4x^3-4x^2-9x+9$4x3−4x2−9x+9 by $\left(x-1\right)$(x−1) 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)=(x−1)(4x2−9).
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)=(x−1)(2x−3)(2x+3).
Example 2:
The function $y=\left(x-2\right)^3-27$y=(x−2)3−27 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 $\left(a-b\right)\left(a^2+ab+b^2\right)$(a−b)(a2+ab+b2), we have:
| $y$y | $=$= | $\left(x-2\right)^3-27$(x−2)3−27 |
| $=$= | $\left[\left(x-2\right)-3\right]\left[\left(x-2\right)^2+3\left(x-2\right)+9\right]$[(x−2)−3][(x−2)2+3(x−2)+9] | |
| $=$= | $\left(x-5\right)\left(x^2-4x+4+3x-6+9\right)$(x−5)(x2−4x+4+3x−6+9) | |
| $=$= | $\left(x-5\right)\left(x^2-x+7\right)$(x−5)(x2−x+7) | |
The quadratic factor is irreducible (the discriminant $\Delta=b^2-4ac=-27$Δ=b2−4ac=−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)=6x3−29x2−7x+10 is known to have the factor $\left(2x-1\right)$(2x−1). 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)=(2x−1)(3x2−13x−10) and the discriminant of the quadratic can be determined as $\Delta=b^2-4ac=49$Δ=b2−4ac=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)=(2x−1)(3x+2)(x−5).
The graph of the function given by $y=\left(2x-1\right)\left(3x+2\right)\left(x-5\right)$y=(2x−1)(3x+2)(x−5) has $x$x-intercepts of $\frac{1}{2},-\frac{2}{3}$12,−23 and $5$5.
More Worked Examples
QUESTION 1
The polynomial $x^3+5x^2+2x-8$x3+5x2+2x−8 has a factor of $x-1$x−1.
By observation, find the quadratic factor of $x^3+5x^2+2x-8$x3+5x2+2x−8.
$x^3+5x^2+2x-8=\left(x-1\right)$x3+5x2+2x−8=(x−1)$($($\editable{}$$)$)
Hence factorise the polynomial completely.
QUESTION 2
The polynomial $4x^3+21x^2+29x+6$4x3+21x2+29x+6 has a factor of $x+3$x+3.
By observation, find the quadratic factor of $4x^3+21x^2+29x+6$4x3+21x2+29x+6.
$4x^3+21x^2+29x+6=\left(x+3\right)$4x3+21x2+29x+6=(x+3)$($($\editable{}$$)$)
Hence factorise the polynomial completely.
QUESTION 3
Consider the expression $x^3-64$x3−64.
Factorise this expression.
Does $x^3-64$x3−64 have any other real-valued linear factors?
Yes
ANo
BState the number of real-valued linear factors that $x^3-64$x3−64 has.