Question

When we have a polynomial as $16{{x}^{4}}+12{{x}^{3}}-10{{x}^{2}}+8x+20$ which is divided by $4x-3$, the quotient and the remainder are, respectively.\[\]
A.$4{{x}^{3}}+6{{x}^{2}}+2x\text{ and }\dfrac{61}{2}$ \[\]
B. $4{{x}^{3}}+6{{x}^{2}}+\dfrac{7}{2}\text{ and }\dfrac{61}{2}$\[\]
C. $6{{x}^{2}}+2x+\dfrac{2}{7}\text{ and }\dfrac{61}{2}$\[\]
D. $4{{x}^{3}}+6{{x}^{2}}+2x+\dfrac{7}{2}\text{and }\dfrac{61}{2}$\[\]

Answer 1

Hint: We have the dividend polynomial $p\left( x \right)=16{{x}^{4}}+12{{x}^{3}}-10{{x}^{2}}+8x+20$ and the divisor polynomial $d\left( x \right)=4x-3$. We use the long division method to divide $p\left( x \right)$ by $d\left( x \right)$ in accordance with Euclidean algorithm where we have to divide until we get remainder $r\left( x \right)$ of degree less than the degree of $d\left( x \right)$.

Complete step-by-step solution
We know that when we divide a divided polynomial $p\left( x \right)$ with degree $n$ by some divisor polynomial $d\left( x \right)$ with degree $m\le n$ then we get the quotient polynomial $q\left( x \right)$ of degree $n-m$ and the remainder polynomial as $r\left( x \right)$ of degree $l < m.$ \[\]
We also know the long division method of polynomials where we divide $p\left( x \right)$ by $d\left( x \right)$ comparing the first terms of the divisor polynomial and the dividend polynomial until we get a remainder $r\left( x \right)$ whose degree is less than the degree of $d\left( x \right)$. We are given in the question the dividend polynomial $p\left( x \right)=16{{x}^{4}}+12{{x}^{3}}-10{{x}^{2}}+8x+20$ and the divisor polynomial $d\left( x \right)=4x-3$. The degree of $p\left( x \right)$ is 4 and the degree of $d\left( x \right)$ is 1. We are going to use the long division method. \[\]
We compare the first term in $p\left( x \right)$ which is $16{{x}^{4}}$ and first term in $d\left( x \right)$ which is $4x$ which is 4. We can only reach $16{{x}^{4}}$ by only multiplying $\dfrac{16{{x}^{4}}}{4x}=4{{x}^{3}}$ to $d\left( x \right)=4x-3$. We initiate the long division method and have,
\[\begin{align}
  & 4x-3\overset{4{x}^{3}}{\overline{\left){16{{x}^{4}}+12{{x}^{3}}-10{{x}^{2}}+8x+20}\right.}} \\
 & \hspace{1.4 cm }\underline{16{{x}^{4}}-12{{x}^{3}}} \\
 & \hspace{2.8 cm }24{{x}^{3}}-10{{x}^{2}}+8x+20 \\
 & \hspace{1 cm } \\
\end{align}\]
We see that the remainder is $24{{x}^{3}}-10{{x}^{2}}+8x+20$. We see that its degree is 3 and larger than degree of $d\left( x \right)$. We know from Euclidean division algorithm this remainder will be new dividend whose firs term is $24{{x}^{3}}$. So we multiply $\dfrac{24{{x}^{3}}}{4x}=16{{x}^{2}}$ to $d\left( x \right)=4x-3$.We continue in the long division method,
\[\begin{align}
& 4x-3\overset{4{{x}^{3}}+6{{x}^{2}}}{\overline{\left){16{{x}^{4}}+12{{x}^{3}}-10{{x}^{2}}+8x+20}\right.}} \\
 & \hspace{1 cm }-\underline{\left( 16{{x}^{4}}-12{{x}^{3}} \right)} \\
 & \hspace{2.8 cm }24{{x}^{3}}-10{{x}^{2}}+8x+20 \\
 & \hspace{2.6 cm }-\underline{\left( 24{{x}^{3}}-18{{x}^{2}} \right)} \\
 & \hspace{3.5 cm }8{{x}^{2}}+8x+20 \\
\end{align}\]
The new remainder is $8{{x}^{2}}+8x+20$ whose degree is 2 larger then a degree of $d\left( x \right)$. So the remainder is our new dividend whose first term is $8{{x}^{2}}$. So we need to multiply $\dfrac{8{{x}^{2}}}{4x}=2x$ to $d\left( x \right)$. We have,
\[\begin{align}
& 4x-3\overset{4{{x}^{3}}+6{{x}^{2}}+2x}{\overline{\left){16{{x}^{4}}+12{{x}^{3}}-10{{x}^{2}}+8x+20}\right.}} \\
 & \hspace{1 cm }-\underline{\left( 16{{x}^{4}}-12{{x}^{3}} \right)} \\
 & \hspace{2.8 cm }24{{x}^{3}}-10{{x}^{2}}+8x+20 \\
 & \hspace{2.6 cm }-\underline{\left( 24{{x}^{3}}-18{{x}^{2}} \right)} \\
 & \hspace{3.5 cm }8{{x}^{2}}+8x+20 \\
 & \hspace{3.5 cm }\underline{-\left( 8{{x}^{2}}-6x \right)} \\
 & \hspace{4 cm }14x+20 \\
\end{align}\]
The new remainder is $14x+20$ whose degree is 1 equal to the degree of $d\left( x \right)$. So the remainder is our new dividend whose first term is $14x$. So we need to multiply $\dfrac{14x}{2x}=\dfrac{7}{2}$ to $d\left( x \right)$. We have,

\[\begin{align}
& 4x-3\overset{4x+6{{x}^{2}}+2x+\dfrac{7}{2}}{\overline{\left){16{{x}^{4}}+12{{x}^{3}}-10{{x}^{2}}+8x+20}\right.}} \\
 & \hspace{1 cm }-\underline{\left( 16{{x}^{4}}-12{{x}^{3}} \right)} \\
 & \hspace{1.5 cm }24{{x}^{3}}-10{{x}^{2}}+8x+20 \\
 & \hspace{1 cm }-\underline{\left( 24{{x}^{3}}-18{{x}^{2}} \right)} \\
 & \hspace{1.5 cm }8{{x}^{2}}+8x+20 \\
 & \hspace{1 cm }\underline{-\left( 8{{x}^{2}}-6x \right)} \\
 & \hspace{1.5 cm }14x+20\hspace{1 cm } \\
 & \hspace{1 cm }\underline{-\left( 14x-\dfrac{21}{2} \right)}\hspace{1 cm } \\
 & \hspace{3 cm }\dfrac{61}{2}\hspace{1 cm } \\
 & \hspace{1 cm } \\
\end{align}\]
So the remainder is $\dfrac{61}{2}$ which is polynomial of degree 0 less that the degree of $d\left( x \right)$. So we stop the division here find the quotient and remainder respectively as
$q\left( x \right)=4{{x}^{3}}+6{{x}^{2}}+2x+\dfrac{7}{2}\text{ and }r \left( x \right)=\dfrac{61}{2}$
So the correct option is D.

Note: We can alternatively use synthetic division method and Ruffini’s rule to find the quotient and the remainder with less calculation. We know from the if factor theorem that $d\left( x \right)$ can be a factor of $p\left( x \right)$ when $r\left( x \right)=0$. So we only need to subtract $\dfrac{61}{2}$ to make $d\left( x \right)=4x-3$ a factor.