Question

Divide the polynomial $3{{x}^{4}}-4{{x}^{3}}-3x-1$ by $x-1$, find its quotient and remainder and hence verify the division algorithm.

Answer 1

Hint: Assume the dividend as $p\left( x \right)$ and the divisor as $g\left( x \right)$. Use the long division method to divide $3{{x}^{4}}-4{{x}^{3}}-3x-1$ by $x-1$. Stop at the step where the constant term is obtained if the polynomial in not fully divisible, which will be the remainder. Finally, verify the division algorithm by using the relation: - dividend = divisor $\times $ quotient + remainder. Solve the R.H.S and if it is equal to the L.H.S then our answers are correct.

Complete step-by-step solution:
Here we have been asked to divide the polynomial $3{{x}^{4}}-4{{x}^{3}}-3x-1$ by $x-1$ and write the quotient and the remainder. Further, we have to verify the division algorithm. Let us assume the dividend as $p\left( x \right)$ and the divisor as $g\left( x \right)$. So we get,
$\Rightarrow p\left( x \right)=3{{x}^{4}}-4{{x}^{3}}-3x-1$ and $g\left( x \right)=x-1$
Dividing $p\left( x \right)$ by $g\left( x \right)$ we get,
$\Rightarrow \dfrac{p\left( x \right)}{g\left( x \right)}=\dfrac{3{{x}^{4}}-4{{x}^{2}}-3x-1}{x-1}$
Using the long division method we get,
\[\Rightarrow \dfrac{p\left( x \right)}{g\left( x \right)}=x-1\overset{3{{x}^{3}}-{{x}^{2}}-x-4}{\overline{\left){\underline{\begin{align}
  & \,\,\,\,\,\,\,3{{x}^{4}}-4{{x}^{2}}-3x-1 \\
 & \underline{\left( - \right)3{{x}^{4}}-3{{x}^{3}}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{x}^{3}}-3x \\
 & \underline{\left( - \right)\,\,\,\,\,\,\,\,\,-{{x}^{3}}+{{x}^{2}}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{x}^{2}}-3x-1 \\
 & \underline{\left( - \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{x}^{2}}+x} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-4x-1 \\
 & \underline{\left( - \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-4x+4} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-5 \\
\end{align}}}\right.}}\]
Now, we cannot divide any further because the term left is having exponent of x less than that of the divisor, so we have the quotient as $3{{x}^{3}}-{{x}^{2}}+x-2$ and the remainder as -3.
Now, the division algorithm states that the dividend, divisor, quotient and remainder must verify the relation: - dividend = divisor $\times $ quotient + remainder. So, substituting the values obtained we get,
\[\begin{align}
  & \Rightarrow 3{{x}^{4}}-4{{x}^{3}}-3x-1=\left( x-1 \right)\times \left( 3{{x}^{3}}-{{x}^{2}}-x-4 \right)+\left( -3 \right) \\
&\Rightarrow3{{x}^{4}}-4{{x}^{3}}-3x-1=3{{x}^{4}}-{{x}^{3}}-{{x}^{2}}-4x-3{{x}^{3}}+{{x}^{2}}+x+4+\left( -5 \right) \\
 & \Rightarrow 3{{x}^{4}}-4{{x}^{3}}-3x-1=3{{x}^{4}}-4{{x}^{3}}-3x-1 \\
\end{align}\]
Therefore, we have L.H.S = R.H.S, hence the division algorithm is verified.

Note: Note that here it is difficult to simplify the dividend into its factors as it is a polynomial of degree 4 that is why we have directly used the long division method. Note that generally we stop at the point where we get the exponent of x in the term of the remainder less than the divisor because we cannot take the negative exponent of x in the quotient. Be careful in calculations and remember the division algorithm.