Question

Divide the polynomial $p\left( x \right)$ by $g\left( x \right)$ and find the quotient and the remainder in each of the following.
 $p\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x-3$, $g\left( x \right)={{x}^{2}}-2$

Answer 1

Hint: We solve this question by first considering the given two polynomials and then dividing the given $p\left( x \right)$ with $g\left( x \right)$. As $p\left( x \right)$ is the dividend we consider the largest degree of x term in $p\left( x \right)$ and see with which should we multiply our divisor, that is $g\left( x \right)$ and then multiply it with $g\left( x \right)$. Then we subtract the obtained polynomial from the dividend and the obtained polynomial becomes our new dividend. We follow the same procedure until we get the degree of x in the dividend less than the degree of x in divisor. Then the remaining term is the remainder and the polynomial with which we have multiplied the divisor is our quotient.

Complete step-by-step answer:
We are asked to divide the polynomial $p\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x-3$ with the polynomial $g\left( x \right)={{x}^{2}}-2$.
We need to find the quotient and remainder for this division.
So, now let us divide the given polynomials.
${{x}^{2}}-2\overset{{}}{\overline{\left){{{x}^{3}}-3{{x}^{2}}+5x-3}\right.}}$
As the first term in the dividend is ${{x}^{3}}$, to get it we need to multiply our divisor with $x$.
So, let us multiply the divisor with x. Then we get,
${{x}^{2}}-2\overset{x}{\overline{\left){\begin{align}
  & {{x}^{3}}-3{{x}^{2}}+5x-3 \\
 & {{x}^{3}}\text{ }-2x \\
\end{align}}\right.}}$
Now let us subtract the obtained value by multiplication from the dividend. Then we get,
${{x}^{2}}-2\overset{x}{\overline{\left){\begin{align}
  & {{x}^{3}}-3{{x}^{2}}+5x-3 \\
 & \underline{{{x}^{3}}\text{ }-2x\text{ }} \\
 & \text{ }-3{{x}^{2}}+7x-3 \\
\end{align}}\right.}}$
Now the new dividend is $-3{{x}^{2}}+7x-3$.
Now the first term in the dividend is $-3{{x}^{2}}$. To get it we need to multiply the divisor with -3. So, let us multiply the divisor with -3. Then we get,
\[{{x}^{2}}-2\overset{x-3}{\overline{\left){\begin{align}
  & {{x}^{3}}-3{{x}^{2}}+5x-3 \\
 & \underline{{{x}^{3}}\text{ }-2x\text{ }} \\
 & \text{ }-3{{x}^{2}}+7x-3 \\
 & \text{ }\underline{-3{{x}^{2}}\text{ }+6} \\
\end{align}}\right.}}\]
Now let us subtract the obtained value from the above obtained dividend. Then we get,
${{x}^{2}}-2\overset{x-3}{\overline{\left){\begin{align}
  & {{x}^{3}}-3{{x}^{2}}+5x-3 \\
 & \underline{{{x}^{3}}\text{ }-2x\text{ }} \\
 & \text{ }-3{{x}^{2}}+7x-3 \\
 & \text{ }\underline{-3{{x}^{2}}\text{ }+6} \\
 & \text{ }7x-9 \\
\end{align}}\right.}}$
Now we need to divide $7x-9$ with ${{x}^{2}}-2$. But the first term in $7x-9$, that is $7x$. We can see that degree of the obtained polynomial, that is $7x-2$, is less than the degree of the divisor. So, we cannot divide it.
So, it is the remainder of the above division.
So, from the above we get the quotient as $x-3$ and the remainder as $7x-9$.
Hence the answer is Quotient $=x-3$, Remainder$=7x-9$.

Note: The common mistake one makes in this type of questions is one might not consider the degree of x when the second term of quotient is multiplied with the polynomial taken and place it wrongly and subtract it, that is one might solve it as,
${{x}^{2}}-2\overset{x-1}{\overline{\left){\begin{align}
  & {{x}^{3}}-3{{x}^{2}}+5x-3 \\
 & \underline{{{x}^{3}}-2{{x}^{2}}\text{ }} \\
 & \text{ }-{{x}^{2}}+5x \\
 & \text{ }\underline{-{{x}^{2}}-2x\text{ }} \\
 & \text{ 7}x-3 \\
\end{align}}\right.}}$
But it is wrong. Here when -2 is multiplied with x, we get $-2x$ not $-2{{x}^{2}}$.