Question

Find the square root of: \[4+25{{x}^{2}}-12x-24{{x}^{3}}+16{{x}^{4}}\]

Answer 1

Hint: In order to find a solution to this problem, we will solve the given polynomial by division method. To find the square root of our polynomial we will first rearrange our polynomial in standard form with descending powers of $x$ and then use division method to find the square root of our polynomial. We will get our solution once we get our remainder as $0$.

Complete step by step solution:
We have our given polynomial as:
\[\Rightarrow 4+25{{x}^{2}}-12x-24{{x}^{3}}+16{{x}^{4}}\]
Now, first we rearrange our polynomial in standard form with descending powers of $x$.
On rearranging, we get:
\[\Rightarrow 16{{x}^{4}}-24{{x}^{3}}+25{{x}^{2}}-12x+4\]
Now, we will start division method:
Now, we will find the square root of the given polynomial \[16{{x}^{4}}-24{{x}^{3}}+25{{x}^{2}}-12x+4\] by following the steps below:
Step 1: First, we will compute the square root of the leading term $\left( 16{{x}^{4}} \right)$ and put it, $\left( 4{{x}^{2}} \right)$, in the two places shown.
Step 2: Then, we will subtract and bring down the next two terms.
Step 3: Next, we will double the currently displayed quotient $\left( 4{{x}^{2}} \right)$ to $8{{x}^{2}}$. Then add a new term, $x$, to the quotient such that the result will remove the first term, $\left( -24{{x}^{3}} \right)$, in the current partial remainder.
Step 4: We will repeat steps $2$ and $3$ until remainder gets $0$.
The division will be as follows:
First, we will compute the square root of the leading term $\left( 16{{x}^{4}} \right)$ and put $\left( 4{{x}^{2}} \right)$ in quotient and divisor and then subtract those terms.
\[\begin{align}
  &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 4{{x}^{2}} \\
 & 4{{x}^{2}}\left| \!{\overline {\,
 \begin{align}
  & 16{{x}^{4}}-24{{x}^{3}}+25{{x}^{2}}-12x+4 \\
 & -16{{x}^{4}} \\
 & \overline{0-24{{x}^{3}}+25{{x}^{2}}-12x+4} \\
\end{align} \,}} \right. \\
\end{align}\]
Next, we will bring down the next two terms and will double the currently displayed quotient $\left( 4{{x}^{2}} \right)$ to $8{{x}^{2}}$. Then add a new term, $x$, to the quotient such that the result will remove the first term, $\left( -24{{x}^{3}} \right)$, in the current partial remainder. Therefore, this becomes:
\[\begin{align}
  & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4{{x}^{2}}-3x \\
 & 8{{x}^{2}}-3x\left| \!{\overline {\,
 \begin{align}
  & -24{{x}^{3}}+25{{x}^{2}}-12x+4 \\
 & \underline{-24{{x}^{3}}+9{{x}^{2}}} \\
 & 0+16{{x}^{2}}-12x+4 \\
\end{align} \,}} \right. \\
\end{align}\]
Next, we will bring down the next two terms and will double the currently displayed quotient $\left( 4{{x}^{2}}-3x \right)$ to $8{{x}^{2}}-6x$ as a divisor. Then add a new term, $2$, to the quotient such that the result will remove the term, $\left( 16{{x}^{2}}-12x+4 \right)$ in the current partial remainder, thus getting remainder $0$.
\[\begin{align}
  & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 4{{x}^{2}}-3x+2 \\
 & 8{{x}^{2}}-6x+2\left| \!{\overline {\,
 \begin{align}
  & 16{{x}^{2}}-12x+4 \\
 & \underline{-16{{x}^{2}}-12x+4} \\
 & 0 \\
\end{align} \,}} \right. \\
\end{align}\]
Hence, the square root of \[16{{x}^{4}}-24{{x}^{3}}+25{{x}^{2}}-12x+4\] is $4{{x}^{2}}-3x+2$.

Note: One can verify if the division method is solved correctly is by multiplying the quotient and the divisor and add the remainder that is by using classic remainder formula,
\[Dividend~=~Quotient\times Divisor +~Remainder\]