Chapter 2.PS, Problem 1PS

To determine

To verify:

The function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ then $f\left(k\right)=r$, where $r=a{k}^3+b{k}^2+ck+d$, using long division

Explanation of Solution

Definition:

The Remainder Theorem:

If a polynomial $f\left(x\right)$ is divided by $x-k$, then the remainder is $r=f\left(k\right)$.

Calculation:

Consider the provided function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$.

Use the remainder theorem $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ is divided by $x-k$ then the provided function is evaluated as,

$\begin{array}{l}\underset{\_}{x-k\begin{array}{c}\hfill a{x}^2+\left(b+ak\right)x+\left(a{k}^2+bk+c\right)\\ \hfill \overline{)\begin{array}{l}a{x}^3+b{x}^2+cx+d\\ a{x}^3-ka{x}^2\downarrow \downarrow \\ (-)(+)\end{array}}\end{array}}\\ \left(b+ak\right)x^2+cx\\ \underset{\_}{\begin{array}{l}\left(b+ak\right)x^2-\left(bk+a{k}^2\right)x\\ (-)(+)\end{array}}\\ \left(a{k}^2+bk+c\right)x+d\\ \underset{\_}{\begin{array}{l}\left(a{k}^2+bk+c\right)x-k\left(a{k}^2+bk+c\right)\\ (-)(+)\end{array}}\\ a{k}^3+b{k}^2+ck+d\\ \end{array}$

Thus, the remainder is $r=a{k}^3+b{k}^2+ck+d$.