Chapter 3: Polynomial Functions

3.4.3: The Remainder Theorem

Learning Objectives

  • Use the remainder theorem to find the reminder of the division of a polynomial divided by a binomial

The Remainder Theorem

Let’s first consider the synthetic division of a polynomial [latex]p(x)=4x^3-x^2-3x+2[/latex] by a linear function [latex]d(x)=x-2[/latex]:

Synthetic division

[latex]\dfrac{4x^3-x^2-3x+2}{x-2}=4x^2+7x+11+\dfrac{24}{x-2}[/latex]

The remainder when we divide by [latex]x-2[/latex] is 24.

Now, let’s evaluate the function value [latex]p(2)[/latex]:

[latex]\begin{aligned}p(x)&=4x^3-x^2-3x+2\\p(2)&=4(2)^3-(2)^2-3(2)+2\\&=32-4-6+2\\&=24\end{aligned}[/latex]

We have just shown that the remainder when we divide by [latex]x-2[/latex] is equal to the function value [latex]p(2)=24[/latex].

This is not a coincidence. If a polynomial is divided by [latex]x–k[/latex], the remainder is always equal to the polynomial function at [latex]k[/latex], that is, [latex]p(k)[/latex]. Let’s walk through the proof of the theorem.

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

[latex]p\left(x\right)=d\left(x\right)q\left(x\right)+r\left(x\right)[/latex]

If the divisor, [latex]d(x)=x-k[/latex], this takes the form

[latex]p\left(x\right)=\left(x-k\right)q\left(x\right)+r(x)[/latex]

Since the divisor [latex]x-k[/latex] has degree 1, the remainder will be a constant, [latex]r[/latex]. And, if we evaluate this for [latex]x=k[/latex], we have

[latex]\begin{cases}\begin{aligned}p(k)&=(k-k)q(k)+r \\ &=0\cdot q\left(k\right)+r \\ &=r\hfill \end{aligned}\end{cases}[/latex]
In other words, [latex]p(k)[/latex] is the remainder obtained by dividing [latex]p(x)[/latex] by [latex]x-k[/latex].

THE REMAINDER THEOREM

For any polynomial [latex]p(x)[/latex] of degree 1 or higher and any linear function [latex]d(x)=x-k[/latex],

[latex]p(k)[/latex] = the remainder when [latex]p(x)[/latex] is divided by [latex]x-k[/latex].

Algebraically:

If [latex]\dfrac{p(x)}{x-k}=q(x)+\dfrac{r}{x-k}[/latex], then [latex]r=p(k)[/latex]

Example 1

Determine the remainder of the division of [latex]f\left(x\right)={x}^{3}-15{x}^{2}+2x - 7[/latex] and [latex]x-2[/latex] without doing the division.

Solution

The remainder theorem tells us that  [latex]\dfrac{f(x)}{d(x)}=\dfrac{{x}^{3}-15{x}^{2}+2x - 7}{x-2}=q(x)+r[/latex] and that [latex]r=f(2)[/latex].

[latex]f\left(2\right)=(2)^{3}-15(2)^{2}+2(2) - 7 = -55[/latex]

So the remainder is –55.

Example 2

Find the remainder of the division of the polynomial [latex]{x}^{2}-2x+3[/latex] and [latex]x+1[/latex].

Solution

Let [latex]p(x)={x}^{2}-2x+3[/latex], then use the remainder theorem to find the remainder = [latex]p(-1)[/latex].

[latex]p(-1)=(-1)^{2}-2(-1)+3 = 1 + 2 + 3 = 6[/latex]

So the remainder is 6.

Try It 1

Determine the remainder when the polynomial [latex]2x^{2}-3x+4[/latex] is divided by [latex]x-3[/latex].

Try It 2

Determine the remainder when the polynomial [latex]-3x^3+2x^{2}-3x+4[/latex] is divided by [latex]x+2[/latex].

 


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