Evaluate a polynomial using the Remainder Theorem

In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem. If the polynomial is divided by x – k, the remainder may be found quickly by evaluating the polynomial function at k, that is, f(k) Let’s walk through the proof of the theorem.

Recall that the Division Algorithm states that, given a polynomial dividend f(x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f(x), there exist unique polynomials q(x) and r(x) such that

f(x)=d(x)q(x)+r(x)f\left(x\right)=d\left(x\right)q\left(x\right)+r\left(x\right)

If the divisor, d(x), is x – k, this takes the form

f(x)=(xk)q(x)+rf\left(x\right)=\left(x-k\right)q\left(x\right)+r

Since the divisor x – k is linear, the remainder will be a constant, r. And, if we evaluate this for xk, we have

{f(k)=(kk)q(k)+r =0q(k)+r =r\begin{cases}f\left(k\right)=\left(k-k\right)q\left(k\right)+r\qquad \\ \text{ }=0\cdot q\left(k\right)+r\qquad \\ \text{ }=r\qquad \end{cases}

In other words, f(k) is the remainder obtained by dividing f(x) by x – k.

A General Note: The Remainder Theorem

If a polynomial

f(x)f\left(x\right)
is divided by x – k, then the remainder is the value
f(k)f\left(k\right)
.

How To: Given a polynomial function
ff
, evaluate
f(x)f\left(x\right)
at
x=kx=k
using the Remainder Theorem.

  1. Use synthetic division to divide the polynomial by
    xkx-k
    .
  2. The remainder is the value
    f(k)f\left(k\right)
    .

Example 1: Using the Remainder Theorem to Evaluate a Polynomial

Use the Remainder Theorem to evaluate

f(x)=6x4x315x2+2x7f\left(x\right)=6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7
 at
x=2x=2
.

Solution

To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by

x2x - 2
.

{2){61152712 221432{ 611 71625\begin{cases}\\ 2\overline{)\begin{cases}6\qquad & -1\qquad & -15\qquad & 2\qquad & -7\qquad \\ \qquad & 12\qquad & \text{ }22\qquad & 14\qquad & 32\qquad \end{cases}}\\ \begin{cases}\text{ }6\qquad & 11\qquad & \text{ }7\qquad & \text{16}\qquad & 25\qquad \end{cases}\end{cases}

The remainder is 25. Therefore,

f(2)=25f\left(2\right)=25
.

Analysis of the Solution

We can check our answer by evaluating

f(2)f\left(2\right)
.

{f(x)=6x4x315x2+2x7f(2)=6(2)4(2)315(2)2+2(2)7=25\begin{cases}f\left(x\right) & =6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7 \\ f\left(2\right) & =6{\left(2\right)}^{4}-{\left(2\right)}^{3}-15{\left(2\right)}^{2}+2\left(2\right)-7 \\ \qquad & =25\qquad \end{cases}

Try It 1

Use the Remainder Theorem to evaluate

f(x)=2x53x49x3+8x2+2f\left(x\right)=2{x}^{5}-3{x}^{4}-9{x}^{3}+8{x}^{2}+2


at
x=3x=-3
.

Solution

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