Question

Divide the polynomial ${{x}^{3}}+4{{x}^{2}}-3x-10$ by $x+1$ and verify your remainder by remainder factor theorem.

Answer 1

Hint: To divide ${{x}^{3}}+4{{x}^{2}}-3x-10$ by $x+1$ , we have to write ${{x}^{3}}+4{{x}^{2}}-3x-10$ as the dividend and $x+1$ as the divisor. Then, we have to divide ${{x}^{3}}$ by x and write the quotient on the top. We have to multiply this quotient with $x+1$ and write the result below the dividend and subtract it from the dividend. We have to perform this operation till we get a remainder. To verify using the remainder theorem, we will equate the divisor to 0 and find the value of x. We have to substitute this value of x in the dividend and check whether the result is equal to the remainder.

Complete step by step answer:
We have to divide the polynomial ${{x}^{3}}+4{{x}^{2}}-3x-10$ by $x+1$ . For this, we will use the long division method. Firstly, we will write ${{x}^{3}}+4{{x}^{2}}-3x-10$ as the dividend and $x+1$ as the divisor.
$x+1\overset{{}}{\overline{\left){{{x}^{3}}+4{{x}^{2}}-3x-10}\right.}}$
Now, we have to divide ${{x}^{3}}$ by x and write the quotient on the top.
\[x+1\overset{\begin{matrix}
   {{x}^{2}} & {} & {} & {} & {} & {} \\
\end{matrix}}{\overline{\left){{{x}^{3}}+4{{x}^{2}}-3x-10}\right.}}\]
We have to multiply the quotient with $x+1$ and write the result below the dividend and subtract it from the dividend.
\[x+1\overset{\begin{matrix}
   {{x}^{2}} & {} & {} & {} & {} & {} \\
\end{matrix}}{\overline{\left){\begin{align}
  & {{x}^{3}}+4{{x}^{2}}-3x-10 \\
 & {{x}^{3}}+{{x}^{2}} \\
 & \begin{matrix}
   - & - \\
\end{matrix} \\
 & \_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
 & \begin{matrix}
   {} & 3{{x}^{2}}-3x-10 \\
\end{matrix} \\
\end{align}}\right.}}\]
Similarly, we have to continue till we get a remainder.

\[x+1\overset{{{x}^{2}}+3x-6}{\overline{\left){\begin{align}
  & {{x}^{3}}+4{{x}^{2}}-3x-10 \\
 & {{x}^{3}}+{{x}^{2}} \\
 & \begin{matrix}
   - & - \\
\end{matrix} \\
 & \_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
 & \begin{matrix}
   {} & 3{{x}^{2}}-3x-10 \\
\end{matrix} \\
 & \begin{matrix}
   {} & 3{{x}^{2}}+3x \\
\end{matrix} \\
 & \begin{matrix}
   {} & - & - \\
\end{matrix} \\
 & \_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
 & \begin{matrix}
   {} & {} & -6x-10 \\
\end{matrix} \\
 & \begin{matrix}
   {} & {} & -6x \\
\end{matrix}-6 \\
 & \begin{matrix}
   {} & {} & + & + \\
\end{matrix} \\
 & \_\_\_\_\_\_\_\_\_\_\_\_\_ \\
 & \begin{matrix}
   {} & {} & {} & -4 \\
\end{matrix} \\
\end{align}}\right.}}\]
Therefore, the quotient when ${{x}^{3}}+4{{x}^{2}}-3x-10$ is divided by $x+1$ is \[{{x}^{2}}+3x-6\] and the remainder is -4.
We have to verify the remainder using the remainder factor theorem. We know that the remainder theorem states that a polynomial p(x) can be divided g(x) only if $p\left( a \right)=0$ , where a is the root of the divisor g(x). Let us consider $p\left( x \right)={{x}^{3}}+4{{x}^{2}}-3x-10$ and $g\left( x \right)=x+1$ . Let us find the root of g(x) by equating it to 0.
$\begin{align}
  & \Rightarrow x+1=0 \\
 & \Rightarrow x=-1 \\
\end{align}$
Let us substitute the above value of x in the polynomial p(x).
$\begin{align}
  & \Rightarrow p\left( -1 \right)={{\left( -1 \right)}^{3}}+4{{\left( -1 \right)}^{2}}-3\left( -1 \right)-10 \\
 & \Rightarrow p\left( -1 \right)=-1+4+3-10 \\
 & \Rightarrow p\left( -1 \right)=-4 \\
\end{align}$
Therefore, the remainder is $p\left( -1 \right)=-4$ .
Hence, verified.

Note: Students must be through with the long division method and the remainder theorem. They must note that before writing the dividend and divisor, they have to arrange the terms of these in the decreasing order of the powers of the exponents.