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