# Remainder Theorem

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The Remainder Theorem enables us to calculate the remainder of the division of any polynomial by a linear polynomial, without actually carrying out the steps of the division algorithm. To see how, we consider the following two polynomials once again:

$\begin{array}{l}a\left(x\right):2x^3-x^2+x-1\\b\left(x\right):x+7\end{array}$

A while back, we calculated (using the division algorithm) that the remainder when $$a\left( x \right)$$ is divided by $$b\left( x \right)$$ will be $$r = - 743$$.

Now, consider what would happen if you were to plugin the zero of $$b\left( x \right)$$ into $$a\left( x \right)$$. The zero of $$b\left( x \right)$$ is $$x = - 7$$. We evaluate $$a\left( x \right)$$ for this value of x:

$\begin{array}{l}a\left(-7\right)=2\left(-7\right)^3-\left(-7\right)^2+\left(-7\right)-1\\\;\;\;\;\;\;\;\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\;-743\end{array}$

Wow! This is the same as the remainder we calculated using the division algorithm. But how did it come out to be the same? Is it a coincidence, or is something else at work here? Let’s look at one more example. Consider the following two polynomials once again:

$\begin{array}{l}a\left(x\right):6x^4-x^3+2x^2-7x+2\\b\left(x\right):2x+3\end{array}$

We calculated the remainder in this case to be $$r = \frac{{203}}{4}$$. Now, let’s see what happens when we evaluate $$a\left( x \right)$$for x equal to the zero of $$b\left( x \right)$$, which is $$x = - \frac{3}{2}$$. We have

\begin{align}&a\left(-\frac32\right)=6\left(-\frac32\right)^4-\left(-\frac32\right)^3+2\left(-\frac32\right)^2-7\left(-\frac32\right)+2\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;=6\left(\frac{81}{16}\right)+\frac{27}8+2\left(\frac94\right)+\frac{21}2+2\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{243}8+\frac{27}8+\frac92+\frac{21}2+2\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{270}8+\frac{30}2+2\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{270+120+16}8\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{406}8\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{203}4\end{align}

Once again, this has turned out to be equal to the remainder we calculated using the division algorithm. There is definitely some link here. To understand that, we consider the general case. Let $$a\left( x \right)$$ be the dividend polynomial and $$b\left( x \right)$$ the linear divisor polynomial, and let $$q\left( x \right)$$ be the quotient and r the constant remainder. Thus, we have

$a\left( x \right) = b\left( x \right)q\left( x \right) + r\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \left( * \right)$

Let us denote the zero of the linear polynomial $$b\left( x \right)$$ by k. This means that

$$b\left( k \right) = 0$$

If we plug-in x as k in the starred relation above, we have

$$a\left( k \right) = b\left( k \right)q\left( k \right) + r$$

Note that doing this is allowed since the starred relation holds true for every value of x. In fact, it is a polynomial identity. Since $$b\left( k \right) = 0,$$ we are left with $$a\left( k \right) = r$$.

In other words, the remainder is equal to the value of $$a\left( x \right)$$ when x is equal to k. Precisely what we stumbled upon!

This is exactly what the Remainder Theorem is: When a polynomial $$a\left( x \right)$$ is divided by a linear polynomial $$b\left( x \right)$$ whose zero is x equal to k, the remainder is given by $$r = a\left( k \right)$$.

Example 1: Find the remainder when

$$p\left( x \right):3{x^5} - {x^4} + {x^3} - 4{x^2} + 2$$

is divided by $$q\left( x \right):x - 1$$.

Solution: We use the Remainder Theorem: we will substitute the zero of $$q\left( x \right)$$ into the polynomial $$p\left( x \right)$$ to find the remainder r:

$\begin{array}{l}r = p\left( 1 \right)\\\,\,\,\, = 3{\left( 1 \right)^5} - {\left( 1 \right)^4} + {\left( 1 \right)^{^3}} - 4{\left( 1 \right)^2} + 2\\\,\,\,\, = 3 - 1 + 1 - 4 + 2\\\,\,\,\, = 1\end{array}$

Example 2: Consider the following polynomial:

$$p\left( x \right):{x^6} - 2{x^5} + {x^4} + 2{x^3} + {x^2} - 36$$

Evaluate the value of this polynomial when x is equal to 2. What can you infer from your answer?

Solution: We have

$\begin{array}{l}p\left( 2 \right) = {\left( 2 \right)^6} - 2{\left( 2 \right)^5} + {\left( 2 \right)^2} + 2{\left( 2 \right)^3} + {2^2} - 36\\\;\;\;\;\;\,\,\, = 64 - 64 + 16 + 16 + 4 - 36\\\;\;\;\;\;\,\,\, = 0\end{array}$

This means that if $$p\left( x \right)$$ is divided by the linear polynomial $$q\left( x \right):x - 2,$$ there will be no remainder! Thus, we can conclude that $$q\left( x \right)$$ is a factor of $$p\left( x \right)$$.

In general, whenever a polynomial is divided by a linear divisor and the remainder is 0, the linear divisor must be a factor of the polynomial.

Polynomials
Grade 9 | Questions Set 1
Polynomials
Polynomials