Question

How do you divide $\dfrac{4{{x}^{4}}-12{{x}^{3}}+5x+3}{\left( {{x}^{2}}+4x+4 \right)}$ ?

Answer 1

Hint: The given polynomial must be divided by using the long division polynomial method. We start to solve the question by dividing the first term of the dividend and the divisor. Then, we multiply the divisor with the answer obtained before. Lastly, we subtract the dividend with the answer obtained in the previous step.
We have been given two polynomials(dividend and a divisor) and need to divide them. We will be using a long division polynomial method to get the required result.

Complete step by step solution:
We need to divide the first term of the dividend with the first term of the divisor.
$\Rightarrow \dfrac{4{{x}^{4}}}{{{x}^{2}}}=4{{x}^{2}}$
Multiplying $4{{x}^{2}}$ with the divisor,
$\Rightarrow 4{{x}^{2}}\left( {{x}^{2}}+4x+4 \right)$
Multiplying $4{{x}^{2}}$ to each term of the expression, we get,
$\Rightarrow 4{{x}^{4}}+16{{x}^{3}}+16{{x}^{2}}$
Subtracting the dividend from the above expression,
$\Rightarrow \left( 4{{x}^{4}}-12{{x}^{3}}+5x+3 \right)-\left( 4{{x}^{4}}+16{{x}^{3}}+16{{x}^{2}} \right)$
Simplifying the above expression, we get,
$\Rightarrow -28{{x}^{3}}-16{{x}^{2}}+5x+3$
The above steps can be represented as follows,
${{x}^{2}}+4x+4\overset{4{{x}^{2}}}{\overline{\left){\begin{align}
  & 4{{x}^{4}}-12{{x}^{3}}+5x+3 \\
 & \left( - \right) \\
 & 4{{x}^{4}}+16{{x}^{3}}+16{{x}^{2}} \\
 & \overline{\left){-28{{x}^{3}}-16{{x}^{2}}+5x+3}\right.} \\
\end{align}}\right.}}$
Now, we need to divide the first term of $-28{{x}^{3}}-16{{x}^{2}}+5x+3$ with the first term of the divisor.
$\Rightarrow \dfrac{-28{{x}^{3}}}{{{x}^{2}}}=-28x$
Multiplying $-28x$ with the divisor,
$\Rightarrow -28x\left( {{x}^{2}}+4x+4 \right)$
Multiplying $-28x$ to each term of the expression, we get,
$\Rightarrow -28{{x}^{3}}-112{{x}^{2}}-112x$
Subtracting $-28{{x}^{3}}-16{{x}^{2}}+5x+3$ and the above expression,
$\Rightarrow \left( -28{{x}^{3}}-16{{x}^{2}}+5x+3 \right)-\left( -28{{x}^{3}}-112{{x}^{2}}-112x \right)$
Simplifying the above expression, we get,
$\Rightarrow 96{{x}^{2}}+117x+3$
The above steps can be represented as follows,
${{x}^{2}}+4x+4\overset{4{{x}^{2}}-28x}{\overline{\left){\begin{align}
  & 4{{x}^{4}}-12{{x}^{3}}+5x+3 \\
 & \left( - \right) \\
 & 4{{x}^{4}}+16{{x}^{3}}+16{{x}^{2}} \\
 & \overline{\left){-28{{x}^{3}}-16{{x}^{2}}+5x+3}\right.} \\
 & \left( - \right) \\
 & -28{{x}^{3}}-112{{x}^{2}}-112x \\
 & \overline{\left){96{{x}^{2}}+117x+3}\right.} \\
\end{align}}\right.}}$
Now, we need to divide the first term of $96{{x}^{2}}+117x+3$ with the first term of the divisor.
$\Rightarrow \dfrac{96{{x}^{2}}}{{{x}^{2}}}=96$
Multiplying 96 with the divisor,
$\Rightarrow 96\left( {{x}^{2}}+4x+4 \right)$
Multiplying $96$ to each term of the expression, we get,
$\Rightarrow 96{{x}^{2}}+384x+384$
Subtracting $96{{x}^{2}}+117x+3$ from the above expression,
$\Rightarrow \left( 96{{x}^{2}}+117x+3 \right)-\left( 96{{x}^{2}}+384x+384 \right)$
Simplifying the above expression, we get,
$\Rightarrow -267x-381$
The above steps can be represented as follows,
${{x}^{2}}+4x+4\overset{4{{x}^{2}}-28x+96}{\overline{\left){\begin{align}
  & 4{{x}^{4}}-12{{x}^{3}}+5x+3 \\
 & \left( - \right) \\
 & 4{{x}^{4}}+16{{x}^{3}}+16{{x}^{2}} \\
 & \overline{\left){-28{{x}^{3}}-16{{x}^{2}}+5x+3}\right.} \\
 & \left( - \right) \\
 & -28{{x}^{3}}-112{{x}^{2}}-112x \\
 & \overline{\left){\begin{align}
  & 96{{x}^{2}}+117x+3 \\
 & \left( - \right) \\
 & 96{{x}^{2}}+384x+384 \\
 & \overline{\left){-267x-381}\right.} \\
\end{align}}\right.} \\
\end{align}}\right.}}$
The degree of the remainder polynomial $\left( -267x-381 \right)$ is 1.
The degree of the divisor polynomial $\left( {{x}^{2}}+4x+4 \right)$ is 2.
As the degree of the remainder polynomial is less than that of the divisor polynomial, the division cannot be further performed.
$\Rightarrow \dfrac{Dividend}{Divisor}=Quotient+\dfrac{\operatorname{Re}mainder}{Divisor}$
$\therefore \dfrac{4{{x}^{4}}-12{{x}^{3}}+5x+3}{\left( {{x}^{2}}+4x+4 \right)}=4{{x}^{2}}-28x+96+\dfrac{\left( -267x-381 \right)}{\left( {{x}^{2}}+4x+4 \right)}$

Note: The Long division method helps us to find the factors of the polynomial. The remainder and quotient of the division can be cross-checked using the formula,$\Rightarrow \dfrac{Dividend}{Divisor}=Quotient+\dfrac{\operatorname{Re}mainder}{Divisor}$