Factorising cubics and quartics

A qua­dratic can always be fac­torised from its zeros using the for­mula. For a cubic or a bi-qua­dratic (quar­tic) poly­no­mial there is no such easy for­mula, so we com­bine two tools we already have. The fac­tor the­o­rem finds one fac­tor; long divi­sion removes it and leaves a poly­no­mial of lower degree; and we repeat until only a qua­dratic is left, which we already know how to fac­torise. This les­son shows the method when a fac­tor is given and when we must find one our­selves.

When one fac­tor is given#

If we are told that $x - a$ (or $ax + b$) is a fac­tor, divide by it. The quo­tient has degree one less, and we fac­torise that.

Exam­ple 1#

Fac­torise $x^3 + 13x^2 + 32x + 20$, given that $x + 2$ is a fac­tor.

Divide by $x + 2$:

$$\begin{aligned} x^3 + 13x^2 + 32x + 20 &= (x + 2)(x^2 + 11x + 10) + 0 \\ &= (x + 2)(x + 1)(x + 10) \end{aligned}$$

The qua­dratic split with prod­uct $10$ and sum $11$: the pair $1$ and $10$.

Exam­ple 2#

Fac­torise $9z^3 - 27z^2 - 100z + 300$, given that $3z + 10$ is a fac­tor.

So $9z^3 - 27z^2 - 100z + 300 = (3z + 10)(3z^2 - 19z + 30)$.

Now fac­torise the qua­dratic. Prod­uct $3 \times 30 = 90$, sum $-19$: the pair $-10$ and $-9$.

$$3z^2 - 10z - 9z + 30 = z(3z - 10) - 3(3z - 10) = (3z - 10)(z - 3)$$

$$9z^3 - 27z^2 - 100z + 300 = (3z + 10)(3z - 10)(z - 3)$$

When no fac­tor is given: trial and error#

Now we must find a fac­tor our­selves. Sub­sti­tute small num­bers $x = 1, -1, 2, -2, \dots$ into $f(x)$ until one gives $0$. By the fac­tor the­o­rem, if $f(a) = 0$ then $x - a$ is a fac­tor.

Which num­bers are worth try­ing? When the lead­ing coef­fi­cient is $1$, any inte­ger zero must divide the con­stant term. For $x^3 - 6x^2 + 3x + 10$ the only can­di­dates are the divi­sors of $10$: $\pm 1, \pm 2, \pm 5, \pm 10$.

Two quick checks save time:

• If the sum of all the coef­fi­cients is $0$, then $f(1) = 0$, so $x - 1$ is a fac­tor.
• If the sum of the coef­fi­cients of the even pow­ers (includ­ing the con­stant term) equals the sum of the coef­fi­cients of the odd pow­ers, then $f(-1) = 0$, so $x + 1$ is a fac­tor.

Exam­ple 3#

Fac­torise $x^3 + 2x^2 - x - 2$.

Let $f(x) = x^3 + 2x^2 - x - 2$. Try $x = 1$: $f(1) = 1 + 2 - 1 - 2 = 0$. So $x - 1$ is a fac­tor.

$$x^3 + 2x^2 - x - 2 = (x - 1)(x^2 + 3x + 2) = (x - 1)(x + 1)(x + 2)$$

Exam­ple 4#

Fac­torise $x^3 - 6x^2 + 3x + 10$.

Let $f(x) = x^3 - 6x^2 + 3x + 10$.

$$f(1) = 1 - 6 + 3 + 10 = 8 \neq 0$$

$$f(-1) = -1 - 6 - 3 + 10 = 0$$

So $x + 1$ is a fac­tor. (The quick check agrees: even pow­ers $-6 + 10 = 4$, odd pow­ers $1 + 3 = 4$.)

$$x^3 - 6x^2 + 3x + 10 = (x + 1)(x^2 - 7x + 10) = (x + 1)(x - 2)(x - 5)$$

Quar­tics: find two fac­tors first#

For a bi-qua­dratic poly­no­mial, look for two zeros. Their two lin­ear fac­tors mul­ti­ply to a qua­dratic fac­tor, and divid­ing by that qua­dratic leaves another qua­dratic.

Exam­ple 5#

Fac­torise $2x^4 + x^3 - 14x^2 - 19x - 6$.

Let $f(x) = 2x^4 + x^3 - 14x^2 - 19x - 6$.

$$f(1) = 2 + 1 - 14 - 19 - 6 = -36 \neq 0$$

$$f(-1) = 2 - 1 - 14 + 19 - 6 = 21 - 21 = 0$$

$$f(2) = 32 + 8 - 56 - 38 - 6 = -60 \neq 0$$

$$f(-2) = 32 - 8 - 56 + 38 - 6 = 70 - 70 = 0$$

So $x + 1$ and $x + 2$ are fac­tors, and so is their prod­uct $(x + 1)(x + 2) = x^2 + 3x + 2$.

The quo­tient $2x^2 - 5x - 3$ splits with prod­uct $-6$ and sum $-5$: $2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)$.

$$2x^4 + x^3 - 14x^2 - 19x - 6 = (x + 1)(x + 2)(2x + 1)(x - 3)$$

Exam­ple 6#

Fac­torise $x^4 - 2x^3 - 7x^2 + 8x + 12$.

Let $f(x) = x^4 - 2x^3 - 7x^2 + 8x + 12$.

$$f(1) = 1 - 2 - 7 + 8 + 12 = 12 \neq 0$$

$$f(-1) = 1 + 2 - 7 - 8 + 12 = 15 - 15 = 0$$

$$f(2) = 16 - 16 - 28 + 16 + 12 = 0$$

So $x + 1$ and $x - 2$ are fac­tors, and $(x + 1)(x - 2) = x^2 - x - 2$ is a fac­tor.

$$\begin{aligned} x^4 - 2x^3 - 7x^2 + 8x + 12 &= (x^2 - x - 2)(x^2 - x - 6) \\ &= (x + 1)(x - 2)(x + 2)(x - 3) \end{aligned}$$

Prac­tice#

1. Fac­torise $x^3 + 13x^2 + 31x - 45$, given that $x + 9$ is a fac­tor.
2. Fac­torise $4x^3 + 20x^2 + 33x + 18$, given that $2x + 3$ is a fac­tor.
3. Fac­torise $x^3 + 6x^2 + 11x + 6$.
4. Fac­torise $x^3 - 6x^2 + 11x - 6$.
5. Fac­torise $3x^3 - x^2 - 3x + 1$.
6. Fac­torise $y^3 - 7y + 6$.
7. Fac­torise $x^3 - 23x^2 + 142x - 120$.
8. Fac­torise $2y^3 - 5y^2 - 19y + 42$.
9. Fac­torise $x^4 + 10x^3 + 35x^2 + 50x + 24$.
10. Fac­torise $x^4 - 7x^3 + 9x^2 + 7x - 10$.
11. Fac­torise $2x^4 - 7x^3 - 13x^2 + 63x - 45$.

Answers#