Factor & Remainder Theorem

Exam code: 4037

Written by: Ashika|Reviewed by: Caroline Carroll|Updated 2 July 2026

Factor Theoremvideo

Factor Theorem

Factor theorem

What is the factor theorem?

The factor theorem is a useful result concerning the roots and factors of polynomials
- In the example below, the polynomial $4 x^{3} + 8 x^{2} - 9 x - 18$ has three (linear) factors
  - $(x + 2), (2 x + 3)$ and $(2 x - 3)$
  - and so it has the three roots $x = - 2, x = - \frac{3}{2}$ and $x = \frac{3}{2}$

For a polynomial $f (x)$ the factor theorem states that:

i) if $f (p) = 0$ , then $(x - p)$ is a factor of $f (x)$
( $x = p$ is a root of $f (x)$ )

and

ii) if $(x - p)$ is a factor of $f (x)$ , then $f (p) = 0$

Worked example

a) Show that $(x - 2)$ is a factor of the polynomial $f (x) = x^{3} + 6 x^{2} - 9 x - 14$ .

(From part (ii) of our definition of factor theorem ...) ... if $(x - 2)$ is a factor of $f (x)$ then $f (2) = 0$ .

$\begin{array}{rcl} f (2) & = & {(2)}^{3} + 6 {(2)}^{2} - 9 (2) - 14 \\ f (2) & = & 8 + 24 - 18 - 14 \\ f (2) & = & 0 \end{array}$

Since $f (2) = 0$ , $(x - 2)$ is factor of $f (x)$ .

b) Use the factor theorem to find another factor of $f (x)$ .

Try $f (1)$ first,

$\begin{array}{rcl} f (1) & = & {(1)}^{3} + 6 {(1)}^{2} - 9 (1) - 14 \\ f (1) & = & 1 + 6 - 9 - 14 \\ f (1) & = & - 16 \end{array}$

Since $f (1) \neq 0$ , $(x - 1)$ is not a factor of $f (x)$ .

Try $f (- 1)$ ,

$\begin{array}{rcl} f (- 1) & = & {(- 1)}^{3} + 6 {(- 1)}^{2} - 9 (- 1) - 14 \\ f (- 1) & = & - 1 + 6 + 9 - 14 \\ f (- 1) & = & 0 \end{array}$

Since $f (- 1) = 0$ , $(x + 1)$ is a factor of $f (x)$ .

$(x + 1)$ is another factor of $f (x)$ .

$(x + 7)$ is the third (linear) factor.
Once one factor is known, polynomial division could be used to find the others. (In this case we were specifically asked to use factor theorem.)

Remainder Theorem

Remainder theorem

What is the remainder theorem?

The factor theorem is actually a special case of the more general remainder theorem
The remainder theorem states that when the polynomial $f (x)$ is divided by $(x - a)$ the remainder is $f (a)$
- You may see this written formally as $f (x) = (x - a) Q (x) + f (a)$
- In polynomial division
  - $Q (x)$ would be the result (at the top) of the division (the quotient)
  - $f (a)$ would be the remainder (at the bottom)
  - $(x - a)$ is called the divisor
- In the case when $f (a) = 0, f (x) = (x - a) Q (x)$ and hence $(x - a)$ is a factor of $f (x)$ – the factor theorem!

How do I solve problems involving the remainder theorem?

If it is the remainder that is of particular interest, the remainder theorem saves the need to carry out polynomial division in full
- e.g. The remainder from ${(x}^{2} - 2 x) \div (x - 3)$ is $3^{2} - 2 \times 3 = 3$
- This is because if $f (x) = x^{2} - 2 x$ and $a = 3$
If the remainder from a polynomial division is known, the remainder theorem can be used to find unknown coefficients in polynomials
- g. The remainder from $(x^{2} + p x) \div (x - 2)$ is 8 so the value of p can be found by solving $2^{2} + p (2) = 8$ , leading to $p = 2$
- In harder problems there may be more than one unknown in which case simultaneous equations would need setting up and solving
The more general version of remainder theorem is if $f (x)$ is divided by $(a x - b)$ then the remainder is $f (\frac{b}{a})$
- The remainder is still found by evaluating the polynomial at the value of $x$ such that $a x - b = 0$ (the divisor is zero) but it is not necessarily an integer

Worked example

The polynomial $p (x)$ is given by $8 x^{4} + a x^{2} + b x - 1$ , where $a$ and $b$ are integer constants.
When $p (x)$ is divided by $(x - 1)$ the remainder is 9.
When $p (x)$ is divided by $(2 x - 1)$ the remainder is 1.
Find the values of $a$ and $b$ .

Remainder theorem: " $f (a)$ is the remainder when $f (x)$ is divided by $(x - a)$ ".

$x - 1 = 0$ when $x = 1$ :

$\begin{array}{rcl} p (1) & = & 9 \\ 8 + a + b - 1 & = & 9 \\ a + b & = & 2 \end{array}$

$2 x - 1 = 0$ when $x = \frac{1}{2}$ :

$\begin{array}{rcl} p (\frac{1}{2}) & = & 1 \\ 8 {(\frac{1}{2})}^{4} + a {(\frac{1}{2})}^{2} + b (\frac{1}{2}) - 1 & = & 1 \\ \frac{1}{2} + \frac{1}{4} a + \frac{1}{2} b - 1 & = & 1 \\ a + 2 b & = & 6 \end{array}$

Solving simultaneously,

$table row blank blank cell stack attributes charalign center stackalign right end attributes row a plus b equals 2 end row row a plus 2 b equals 6 end row horizontal line row b equals 4 end row end stack end cell end table$

$∴ a = 2 - 4 = - 2$

$a = - 2, b = 4$

$p (x) = 8 x^{4} - 2 x^{2} + 4 x - 1$