Question

How do you divide $\dfrac{4{{x}^{4}}-12{{x}^{2}}-x-20}{{{x}^{2}}+4}$ ?

Answer 1

Hint: Now to solve the given problem we will first consider the numerator. Now we will rewrite the terms of the numerator such that we can take ${{x}^{2}}+4$ out of the equation. Then we can rewrite the expression as $\text{dividend = divisor}\times \text{quotient + remainder}$ . Hence using this we can find the solution of the given problem.

Complete step by step solution:
Now to solve the given problem we will first simplify the numerator. Now we will simplify the numerator such that we can take ${{x}^{2}}+4$ common out of the terms. Hence we get,
Now let us consider the numerator of the given expression \[4{{x}^{4}}-12{{x}^{2}}-x-20\]
Since we want to take ${{x}^{2}}+4$ common we will first rewrite the term $-12{{x}^{2}}$ as $16{{x}^{2}}-28{{x}^{2}}$
Hence we get the expression as $4{{x}^{4}}+16{{x}^{2}}-28{{x}^{2}}-x-20$
Now simplifying the terms we get, $4{{x}^{2}}\left( {{x}^{2}}+4 \right)-28{{x}^{2}}-x-20$
Now again we will write – 20 as $4{{x}^{2}}\left( {{x}^{2}}+4 \right)-28{{x}^{2}}-112+112-x-20$
Now again simplifying the expression by taking 28 common we get,
$\Rightarrow 4{{x}^{2}}\left( {{x}^{2}}+4 \right)-28\left( {{x}^{2}}+4 \right)-x+92$
Now we know that we cannot take ${{x}^{2}}+4$ common from the term $-x+92$ . Hence we get,
$\Rightarrow 4{{x}^{4}}-12{{x}^{2}}-x-20=\left( {{x}^{2}}+4 \right)\left( 4{{x}^{2}}-28 \right)-x+92$
Now comparing the RHS of the equation with $aq+r$ where a is the divisor, q is the quotient and r is the remainder we get,
Quotient is $\left( 4{{x}^{2}}-28 \right)$ and the remainder is $\left( -x+92 \right)$ .
Hence the solution of $\dfrac{4{{x}^{4}}-12{{x}^{2}}-x-20}{{{x}^{2}}+4}$ is $\left( 4{{x}^{2}}-28 \right)$ and the remainder is $-x+92$.

Note: Now note that we can also use the normal division of polynomials to find the solution of the given division. Note that we continue to divide till we get a polynomial of degree less than the degree of polynomial in divisor.