Remainder Theorem

The remainder theorem is a theorem that looks incredibly simple but has profound time-saving implications in the context of the analysis of polynomials and their factors. It also typically appears in at least one question on the SAT exam.

Remainder Theorem

Let $p(x)$ be a polynomial and $a$ be a constant. The remainder of the division $p(x)÷(x-a)$ is given by $p(a).$

Why?

To create an analogy, recall the way that you computed quotients and remainders in elementary school. For example, $13÷3=4\text{ R}1$ (the quotient of $13÷3$ is $4$ with a remainder of $1$). The reason why this is the correct result is because

$$13=4\cdot 3+1$$

which can be thought of as

$$\text{numerator}=(\text{quotient})\cdot (\text{denominator})+\text{remainder}.$$

All divisions can be formatted in this way, including polynomial divisions. So, in our context here, the quotient $q(x)$ and remainder $r$ of the division $p(x)÷(x-a)$ is related to $p(x)$ and $x-a$ as follows.

$$p(x)=q(x)\cdot (x-a)+r$$

Plugging in $x=a$ gives $p(a)=r,$ proving the theorem.

The following is an important consequence of the remainder theorem.

Note

Let $p(x)$ be a polynomial and $a$ be a constant. If $p(a)=0$, then $x-a$ is a factor of $p(x)$.

Why?

According to the remainder theorem, the remainder of the division $p(x)÷(x-a)$ is $p(a)=0$. Since the remainder of the division is zero, we must have that $x-a$ is a factor of $p(x)$.

Practice Problems

Do not use polynomial division to solve any of the following problems (the remainder theorem will be much, much quicker).

Problem 1

What is the remainder of $x^{10}+x-1$ when divided by $x+1$?

Solution 1

Let $p(x)=x^{10}+x-1.$ Since $x+1=x-(-1),$ the remainder is

$$p(-1)=(-1{)}^{10}+(-1)-1=1-1-1=\boxed{-1}.$$

Problem 2

The remainder when $f(x)$ is divided by $x-2$ is $7$. What is $f(2)$?

Solution 2

By the remainder theorem, $f(2)$ is the remainder of the division $f(x)÷(x-2)$, so clearly $f(2)=\boxed{7}.$

Problem 3

If $f(x)=2x^3+k{x}^2-5x+4$ has a factor $x+3$, what is the value of $k$?

Solution 3

If $x+3$ is a factor of $f(x)$, then we must have that $f(-3)=0.$

$$\begin{array}{c}f(-3)=0\\ 2(-3{)}^3+k(-3{)}^2-5(-3)+4=0\\ 9k-35=0\\ \boxed{k=\frac{35}{9}}\end{array}$$

Problem 4

When $f(x)$ is divided by $x-5,$ the remainder is $0.$ Which of the following must be true?

(A) $f(0)=5$

(B) $f(5)=0$

(C) $f(5)=5$

(D) $f(x)$ has degree $1$

Solution 4

The correct answer is $\boxed{(B)f(5)=0}.$ By the remainder theorem, the remainder of the division $f(x)÷(x-5)$ is $f(5)$, which is known to be zero.

Problem 5

Determine the remainder of $x^4-2x+1$ when divided by each of the following.

• $x-1$
• $x+1$
• $x$

Solution 5

• The remainder of $(x^4-2x+1)÷(x-1)$ is $1^4-2\cdot 1+1=1-2+1=\boxed{0}.$
• The remainder of $(x^4-2x+1)÷(x+1)$ is $(-1{)}^4-2(-1)+1=1+2+1=\boxed{4}.$
• The remainder of $(x^4-2x+1)÷x$ is $0^4-2\cdot 0+1=\boxed{1}$.

Problem 6

Determine the value of $k$ such that $x-5$ divides $f(x)=x^2+kx+20$ and then factor $f(x)$ completely.

Solution 6

If $x-5$ divides $f(x)$, then the remainder of $f(x)÷5$ is zero. By the remainder theorem, we then have

$$\begin{array}{c}f(5)=0\\ 5^2+k\cdot 5+20=0\\ 5k+45=0\\ \boxed{k=-9}\end{array}$$

So, $f(x)=x^2-9x+20=\boxed{(x-4)(x-5)}.$

Problem 7

Determine the values of $k$ and $m$ such that both $x-3$ and $x+1$ divide $f(x)=x^3+m{x}^2+kx+27$ and then factor $f(x)$ completely.

Solution 7

Since $f(x)÷(x-3)$ and $f(x)÷(x+1)$ both have a remainder of zero, we then have $f(3)=0$ and $f(-1)=0$. Let’s determine the consequences of each restriction through some algebraic simplification.

$$\begin{array}{cc}f(3)=0& f(-1)=0\\ 3^3+m\cdot 3^2+k\cdot 3+27=0& (-1{)}^3+m(-1{)}^2+k(-1)+27=0\\ 27+9m+3k+27=0& -1+m-k+27=0\\ 9m+3k+54=0& m-k+26=0\end{array}$$

We now have a system of equations.

$$\{\begin{array}{l}9m+3k+54=0\\ m-k+26=0\end{array}$$

The solution to this system of equations is $\boxed{m=-11}$ and $\boxed{k=15}$.

$$f(x)=x^3-11x^2+15x+27=\boxed{(x-3)(x+1)(x-9)}$$

Problem 8

The division $(x^3+2x-1)÷(x+2)$ yields a quotient of $x^2-2x+6$ Determine the value of the real number $z$ to make the following identity true.

$$x^3+2x-1=x^2-2x+6+\frac{z}{x+2}$$

Solution 8

Since $\frac{\text{numerator}}{\text{denominator}}=\text{quotient}+\frac{\text{remainder}}{\text{denominator}},$ the number $z$ is the remainder of the division $(x^3+2x-1)÷(x+2)$, so $z=(-2{)}^3+2(-2)-1=\boxed{-13}.$