Question

Divide the polynomial function $5{{x}^{3}}-13{{x}^{2}}+21x-14$ by $3-2x+{{x}^{2}}$ and verify the division algorithm.

Answer 1

Hint: First of all rearrange the divisor $3-2x+{{x}^{2}}$ by taking the term containing the highest exponent of x in the first place and writing the next terms according to the exponent of x decreasing. Now, use the long division method to divide $5{{x}^{3}}-13{{x}^{2}}+21x-14$ by ${{x}^{2}}-2x+3$. Stop at the step where the linear expression in x is obtained if the polynomial is 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 $5{{x}^{3}}-13{{x}^{2}}+21x-14$ by $3-2x+{{x}^{2}}$ and 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,
$\Rightarrow p\left( x \right)=5{{x}^{3}}-13{{x}^{2}}+21x-14$ and $g\left( x \right)=3-2x+{{x}^{2}}$
Rearranging the terms of $g\left( x \right)$ by writing them according to the decreasing exponent of x we get,
$\Rightarrow g\left( x \right)={{x}^{2}}-2x+3$
Dividing $p\left( x \right)$ by $g\left( x \right)$ we get,
$\Rightarrow \dfrac{p\left( x \right)}{g\left( x \right)}=\dfrac{5{{x}^{3}}-13{{x}^{2}}+21x-14}{{{x}^{2}}-2x+3}$
Using the long division method we get,
\[\Rightarrow \dfrac{p\left( x \right)}{g\left( x \right)}={{x}^{2}}-2x+3\overset{5x-3}{\overline{\left){\underline{\begin{align}
  & \,\,\,\,\,\,\,5{{x}^{3}}-13{{x}^{2}}+21x-14 \\
 & \underline{\left( - \right)5{{x}^{3}}-10{{x}^{2}}+15x} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3{{x}^{2}}+6x-14 \\
 & \underline{\left( - \right)\,\,\,\,\,\,\,\,\,\,\,\,-3{{x}^{2}}+6x-9} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-5 \\
\end{align}}}\right.}}\]
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 $\left( 5x-3 \right)$ and the remainder as -5.
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,
\[\begin{align}
  & \Rightarrow 5{{x}^{3}}-13{{x}^{2}}+21x-14=\left( {{x}^{2}}-2x+3 \right)\times \left( 5x-3 \right)+\left( -5 \right) \\
 & \Rightarrow 5{{x}^{3}}-13{{x}^{2}}+21x-14=5{{x}^{3}}-3{{x}^{2}}-10{{x}^{2}}+6x+15x-9+\left( -5 \right) \\
 & \Rightarrow 5{{x}^{3}}-13{{x}^{2}}+21x-14=5{{x}^{3}}-13{{x}^{2}}+21x-14 \\
\end{align}\]
Therefore, we have L.H.S = R.H.S, hence the division algorithm is verified.

Note: Here it is difficult to simplify the dividend into its factors 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. You must remember the division algorithm.