Question 2: Find the square root of the following polynomials by division method:
\((iv)\) \(\; \)\(4x^4-12x^3+37x^2-42x+49 \)
Solution:
$\begin{array}{ r l }
& \ \ \ \ \ \underline{\ \ \ \ \ \ 2x^{2} -3x+7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\
& 2x^{2}\bigr) \ \ \cancel{4x^{4}} -12x^{3} +37x^{2} -42x+49\\
& \ \ \ \ \ \ \ \ \underline{\underset{-}{+}\cancel{4x^{4}} \ \ \ \ \ \ \ \ \ \ }\\
& 4x^{2} -3x\ \overline{) -\cancel{12x^{3}} +37x^{2} -42x+49}\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{\underset{+}{-} \ \cancel{12x^{3}} \ \underset{-}{+} \ 9x^{2} \ \ \ \ } \ \\
& 4x^{2} -6x+7\ \overline{) \ \ 28x^{2} -42x+49}\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{\underset{+}{-} 28x^{2} \ \underset{-}{+} 42x \ \underset{-}{+} \ 49\ }\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0
\end{array}$
So, $\displaystyle \text{Square Root } =\pm \left( 2x^{2} -3x+7\right)$