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.
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.
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.
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.
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).
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.
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.
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{}$$)$)
Hence factorise the polynomial completely.
The polynomial $4x^3+21x^2+29x+6$4x3+21x2+29x+6 has a factor of $x+3$x+3.
By observation, find the quadratic factor of $4x^3+21x^2+29x+6$4x3+21x2+29x+6.
$4x^3+21x^2+29x+6=\left(x+3\right)$4x3+21x2+29x+6=(x+3)$($($\editable{}$$)$)
Hence factorise the polynomial completely.
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
No
State the number of real-valued linear factors that $x^3-64$x3−64 has.