The factor theorem

The remain­der the­o­rem tells us the remain­der on divid­ing by $x - a$. The most inter­est­ing case is when that remain­der is zero, because then $x - a$ divides the poly­no­mial exactly: it is a fac­tor. This spe­cial case is called the fac­tor the­o­rem. It lets us test a fac­tor by sub­sti­tut­ing one num­ber, find unknown coef­fi­cients, prove divis­i­bil­ity with­out divid­ing, and, in the next lessons, fac­torise cubics and quar­tics.

The the­o­rem#

Fac­tor the­o­rem. $x - a$ is a fac­tor of $f(x)$ if and only if $f(a) = 0$.

Why it is true. By the remain­der the­o­rem, divid­ing $f(x)$ by $x - a$ leaves the remain­der $f(a)$. If $f(a) = 0$, the remain­der is $0$, so $f(x) = (x - a)\,q(x)$ and $x - a$ is a fac­tor. Going the other way, if $x - a$ is a fac­tor, the remain­der is $0$, so $f(a) = 0$.

As with the remain­der the­o­rem, a lin­ear divi­sor with a coef­fi­cient fol­lows the same pat­tern:

$$\displaystyle ax - b \text{ is a factor of } f(x) \iff f\left(\frac{b}{a}\right) = 0, \qquad ax + b \text{ is a factor of } f(x) \iff f\left(-\frac{b}{a}\right) = 0$$

Exam­ple 1#

Check whether $x - 2$ is a fac­tor of $x^2 - 4$.

$$f(2) = 2^2 - 4 = 4 - 4 = 0$$

So $x - 2$ is a fac­tor. Indeed $x^2 - 4 = (x - 2)(x + 2)$.

Exam­ple 2#

Show that $x - 3$ is a fac­tor of $x^3 - 3x^2 + 4x - 12$.

$$f(3) = 27 - 27 + 12 - 12 = 0$$

So $x - 3$ is a fac­tor.

Find­ing unknown coef­fi­cients#

Exam­ple 3#

Find $a$ so that $2x^4 - ax^3 + 4x^2 + 2x + 1$ is divis­i­ble by $1 - 2x$.

$1 - 2x = 0$ gives $\displaystyle x = \frac{1}{2}$. Divis­i­ble means the remain­der is $0$, so $\displaystyle f\left(\frac{1}{2}\right) = 0$.

$$\displaystyle \begin{aligned} 2\left(\tfrac{1}{16}\right) - a\left(\tfrac{1}{8}\right) + 4\left(\tfrac{1}{4}\right) + 2\left(\tfrac{1}{2}\right) + 1 &= 0 \\ \tfrac{1}{8} - \tfrac{a}{8} + 1 + 1 + 1 &= 0 \\ \tfrac{1 - a}{8} &= -3 \\ 1 - a &= -24 \\ a &= 25 \end{aligned}$$

Exam­ple 4#

The poly­no­mial $ax^3 + bx^2 + x - 6$ has $x + 2$ as a fac­tor and leaves a remain­der $4$ when divided by $x - 2$. Find $a$ and $b$.

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

Since $x + 2$ is a fac­tor, $f(-2) = 0$:

$$-8a + 4b - 2 - 6 = 0 \implies -8a + 4b = 8 \implies -2a + b = 2 \quad (1)$$

Since the remain­der on divid­ing by $x - 2$ is $4$, $f(2) = 4$:

$$8a + 4b + 2 - 6 = 4 \implies 8a + 4b = 8 \implies 2a + b = 2 \quad (2)$$

Sub­tract (1) from (2): $4a = 0$, so $a = 0$. Then from (2), $b = 2$.

So $a = 0$ and $b = 2$, and the poly­no­mial is really $2x^2 + x - 6$. Check: $f(-2) = 8 - 2 - 6 = 0$ and $f(2) = 8 + 2 - 6 = 4$. Both con­di­tions hold.

Exam­ple 5#

Find $a$ and $b$ so that $x + 1$ and $x - 1$ are both fac­tors of $f(x) = x^4 + ax^3 - 3x^2 + 2x + b$.

$$f(-1) = 0 \implies 1 - a - 3 - 2 + b = 0 \implies -a + b = 4 \quad (1)$$

$$f(1) = 0 \implies 1 + a - 3 + 2 + b = 0 \implies a + b = 0 \quad (2)$$

Adding: $2b = 4$, so $b = 2$, and then $a = -2$.

Exam­ple 6#

Find $p$ and $q$ so that $f(x) = x^4 + px^3 + 2x^2 - 3x + q$ is divis­i­ble by $x^2 - 1$.

Since $x^2 - 1 = (x + 1)(x - 1)$, divis­i­bil­ity by $x^2 - 1$ means both $x + 1$ and $x - 1$ are fac­tors.

$$f(-1) = 0 \implies 1 - p + 2 + 3 + q = 0 \implies -p + q = -6 \quad (1)$$

$$f(1) = 0 \implies 1 + p + 2 - 3 + q = 0 \implies p + q = 0 \quad (2)$$

Adding: $2q = -6$, so $q = -3$, and then $p = 3$.

Divis­i­bil­ity with­out actual divi­sion#

To show a poly­no­mial is divis­i­ble by a qua­dratic, fac­torise the qua­dratic into two lin­ear fac­tors and test each one.

Exam­ple 7#

With­out actual divi­sion, prove that $2x^4 - 5x^3 + 2x^2 - x + 2$ is exactly divis­i­ble by $x^2 - 3x + 2$.

First fac­torise the divi­sor: $x^2 - 3x + 2 = x^2 - x - 2x + 2 = (x - 1)(x - 2)$.

Let $f(x) = 2x^4 - 5x^3 + 2x^2 - x + 2$.

$$f(1) = 2 - 5 + 2 - 1 + 2 = 6 - 6 = 0$$

$$f(2) = 2(16) - 5(8) + 2(4) - 2 + 2 = 32 - 40 + 8 = 0$$

So $x - 1$ and $x - 2$ are both fac­tors, and there­fore their prod­uct $x^2 - 3x + 2$ is a fac­tor of $f(x)$.

Mak­ing a poly­no­mial exactly divis­i­ble#

By the divi­sion algo­rithm, $f(x) = g(x)q(x) + r(x)$. Rear­rang­ing,

$$f(x) - r(x) = g(x)\,q(x)$$

so $f(x) - r(x)$ is exactly divis­i­ble by $g(x)$. In words: sub­tract the remain­der, or equiv­a­lently add the remain­der with its sign changed.

Exam­ple 8#

What must be sub­tracted from $4x^4 - 2x^3 - 6x^2 + x - 5$ so that the result is exactly divis­i­ble by $2x^2 + x - 1$?

We divided exactly this pair in the les­son on long divi­sion: the quo­tient was $2x^2 - 2x - 1$ and the remain­der was $-6$. So we must sub­tract $-6$. (Sub­tract­ing $-6$ is the same as adding $6$, giv­ing $4x^4 - 2x^3 - 6x^2 + x + 1$.)

Exam­ple 9#

What must be added to $x^4 + 2x^3 - 2x^2 + x - 1$ so that the result is exactly divis­i­ble by $x^2 + 2x - 3$?

The remain­der is $-x + 2$. To be added: $-(-x + 2) = x - 2$.

A fac­tor for every odd power#

Show that $x + a$ is a fac­tor of $x^n + a^n$ when­ever $n$ is an odd pos­i­tive inte­ger.

Let $f(x) = x^n + a^n$. Then

$$f(-a) = (-a)^n + a^n = -a^n + a^n = 0$$

since an odd power of a neg­a­tive num­ber is neg­a­tive. So $x + a$ is a fac­tor. For exam­ple, $x + 2$ is a fac­tor of $x^3 + 8 = x^3 + 2^3$, and indeed $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$; like­wise $x + 2$ is a fac­tor of $x^5 + 32 = x^5 + 2^5$. The last les­son of this chap­ter devel­ops this idea fully.

Prac­tice#

1. Is $x - 3$ a fac­tor of $x^3 - 6x^2 + 11x - 6$?
2. Is $x + 5$ a fac­tor of $3x^4 + 17x^3 + 9x^2 - 7x - 10$?
3. Is $3x - 2$ a fac­tor of $3x^3 + x^2 - 20x + 12$?
4. For what value of $a$ is $x - 5$ a fac­tor of $x^3 - 3x^2 + ax - 10$?
5. Find $a$ if $x + 2$ is a fac­tor of $4x^4 + 2x^3 - 3x^2 + 8x + 5a$.
6. Find $a$ and $b$ so that $x^3 + 10x^2 + ax + b$ is exactly divis­i­ble by both $x - 1$ and $x - 2$.
7. Find $\alpha$ and $\beta$ if $x + 1$ and $x + 2$ are fac­tors of $x^3 + 3x^2 - 2\alpha x + \beta$.
8. Find $a$ and $b$ if $x^2 - 4$ is a fac­tor of $ax^4 + 2x^3 - 3x^2 + bx - 4$.
9. With­out actual divi­sion, prove that $2x^4 - 6x^3 + 3x^2 + 3x - 2$ is exactly divis­i­ble by $x^2 - 3x + 2$.
10. What must be added to $x^3 - 3x^2 - 12x + 19$ so that the result is exactly divis­i­ble by $x^2 + x - 6$?
11. What must be sub­tracted from $x^3 - 6x^2 - 15x + 80$ so that the result is exactly divis­i­ble by $x^2 + x - 12$?

Answers#