Question

How do you divide $\dfrac{8{{x}^{3}}+5{{x}^{2}}-12x+10}{{{x}^{2}}-3}$ ?

Answer 1

Hint: To divide the cubic equation by quadratic equation, divide the highest order term in the dividend by the highest order term in the divisor. Then multiply the new quotient term by the divisor and solve it by the elimination method. Repeat the steps until you get the remainder. Finally, we will get the quotient and remainder.

Complete step by step answer:
From the given question we have $\dfrac{8{{x}^{3}}+5{{x}^{2}}-12x+10}{{{x}^{2}}-3}$ .
For solving the cubic equation with the quadratic equation we need to follow the steps one by one.
Firstly, divide the highest order term in the dividend 8${{x}^{3}}$ by the highest order term in divisor ${{x}^{2}}$we can get
\[{{x}^{2}}+0x-3\overset{8x}{\overline{\left){8{{x}^{3}}+5{{x}^{2}}-12x+10}\right.}}\]
If there is not term insert zero in that place so we can write like this in the place of x ‘0x’.
Now, multiply the new quotient by the divisor.
We get $8{{x}^{3}}+0{{x}^{2}}-24x$
\[{{x}^{2}}+0x-3\overset{8x}{\overline{\left){\begin{align}
  & 8{{x}^{3}}+5{{x}^{2}}-12x+10 \\
 & 8{{x}^{3}}+0{{x}^{2}}-24x \\
\end{align}}\right.}}\]
Change the sign and subtract it from the dividend, which is $5{{x}^{2}}+12x$ .
\[{{x}^{2}}+0x-3\overset{8x}{\overline{\left){\begin{align}
  & 8{{x}^{3}}+5{{x}^{2}}-12x+10 \\
 & \underline{8{{x}^{3}}+0{{x}^{2}}-24x} \\
 & 0{{x}^{3}}+5{{x}^{2}}+12x \\
\end{align}}\right.}}\]
Again, divide the highest order term in the dividend $5{{x}^{2}}$ by the highest order term in the divisor ${{x}^{2}}$.
So, we will get the quotient as 8x+5.
Multiply the new quotient 5 by the divisor which is $5{{x}^{2}}+\text{ }0x\text{ }-15$. Then change the sign and do the subtraction.
\[{{x}^{2}}+0x-3\overset{8x+5}{\overline{\left){\begin{align}
  & 8{{x}^{3}}+5{{x}^{2}}-12x+10 \\
 & \underline{8{{x}^{3}}+0{{x}^{2}}-24x} \\
 & 0{{x}^{3}}+5{{x}^{2}}+12x \\
 & 5{{x}^{2}}+0x-15 \\
\end{align}}\right.}}\]
Then change the sign and do the subtraction.
\[{{x}^{2}}+0x-3\overset{8x+5}{\overline{\left){\begin{align}
  & 8{{x}^{3}}+5{{x}^{2}}-12x+10 \\
 & \underline{{}_{\left( - \right)}\left( 8{{x}^{3}}+0{{x}^{2}}-24x \right)} \\
 & 0{{x}^{3}}+5{{x}^{2}}+12x \\
 & \underline{{}_{\left( - \right)}\left( 5{{x}^{2}}+0x-15 \right)} \\
 & 12x+25 \\
\end{align}}\right.}}\]
The resulting remainder is 12x+25.
Finally, from the division, we get the quotient as 8x+5 and the remainder as 12x+25.

∴ The answer is 8x+5.

Note: Division of dividend by divisor and multiplication with divisor is important. So, while dividing and multiplying check the coefficient and the powers properly. Some facts of division are: the product of divisor and the quotient added to the remainder is always equal to the dividend. In division, the remainder is always smaller than the divisor. The dividend is always a multiple of the quotient and divisor if there is no remainder.