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. 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:
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.
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.
Sum of two cubes: $a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$a3+b3=(a+b)(a2−ab+b2)
Difference of two cubes: $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$a3−b3=(a−b)(a2+ab+b2)
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!
Factorise $x^3-1000$x3−1000.
Simplify $\frac{125x^3+8}{5x+2}$125x3+85x+2.
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.
Factorise the following expression:
$17x^3+5x^2+17x+5$17x3+5x2+17x+5
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.
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.
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.
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.
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.
Consider the division $\left(x^2-6x+1\right)\div\left(x+3\right)$(x2−6x+1)÷(x+3).
What term needs to be brought down to move onto the next step in the algorithm?
$x$x | |||||||
$x$x | $+$+ | $3$3 | $x^2$x2 | $-$− | $6x$6x | $+$+ | $1$1 |
$x^2$x2 | $+$+ | $3x$3x | |||||
$-$− | $9x$9x |
What is the remainder of this division?
$x$x | $-$− | $9$9 | |||||
$x$x | $+$+ | $3$3 | $x^2$x2 | $-$− | $6x$6x | $+$+ | $1$1 |
$x^2$x2 | $+$+ | $3x$3x | |||||
$-$− | $9x$9x | $+$+ | $1$1 | ||||
$-$− | $9x$9x | $-$− | $27$27 |
Without including the remainder, what is the quotient of this division?
Rewrite $x^2-6x+1$x2−6x+1 in terms of the divisor, the quotient and the remainder.
$x^2-6x+1$x2−6x+1$=$=$\left(x+\editable{}\right)\left(x-\editable{}\right)+\editable{}$(x+)(x−)+
Consider the following division:
$\left(3x^3-15x^2+2x-10\right)\div\left(x-5\right)$(3x3−15x2+2x−10)÷(x−5)
Fill in the gaps to complete the long division process below.
$\editable{}$ $+$+ | $0x+$0x+ | $\editable{}$ | ||
$x-5$x−5 | $3x^3$3x3 | $-$−$15x^2$15x2 | $+$+$2x$2x | $-$−$10$10 |
$\editable{}$ | ||||
$\editable{}$ | ||||
$\editable{}$ | ||||
$\editable{}$ |
State the quotient and remainder when $3x^3-15x^2+2x-10$3x3−15x2+2x−10 is divided by $x-5$x−5.
Quotient = $\editable{}$
Remainder = $\editable{}$
So far we have been given a linear factor but how could we find one if it wasn't given?
We could use technology or the remainder theorem.
The remainder theorem tells use that if we divide a polynomial by $\left(x-r\right)$(x−r), the remainder will be the polynomial evaluated at $x=r$x=r.
Remember that if we divide a polynomial by a factor we will get a remainder of zero. This gives us a great way to find factors using a guess and check method:
Good candidates for guesses for 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 tells us 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, we could try factors of $6$6 :$\left\{1,-1,2,-2,3,-3,6,-6\right\}${1,−1,2,−2,3,−3,6,−6}. Can you find a factor?
It may not be possible to find some linear factors quickly using this method.
Consider the cubic $x^3+9x^2+26x+24$x3+9x2+26x+24.
State one linear factor of the cubic.
Hence factorise the cubic.
When $3x^3-2x^2-4x+k$3x3−2x2−4x+k is divided by $x-3$x−3, the remainder is $47$47. Find the value of $k$k.