Question 2: Find HCF of the following expressions by using division method:
\((iv)\) \(\; \) \(2x^3-4x^2+6x,x^3-2x,3x^2-6x \)
Solution:
$\displaystyle 2x^{3} -4x^{2} +6x=2\left( x^{3} -2x^{2} +3x\right)$
$\displaystyle x^{3} -2x$
$\displaystyle 3x^{2} -6x=3\left( x^{2} -2x\right)$
$\displaystyle 2$ and $\displaystyle 3$ are not common in given polynomials, so we ignore them and conisder only
$\displaystyle x^{3} -2x^{2} +3x$
$\displaystyle x^{3} -2x$
$\displaystyle x^{2} -2x$
$\begin{array}{ r l }
& \ \underline{\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\
x^{2} -2x & \bigr) \ x^{3} -2x^{2} +3x\\
& \underline{\underset{\ \ -}{} x^{3} \ \underset{+}{-}2x^{2} \ \ \ \ \ \ \ \ \ }\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x
\end{array}$
$\displaystyle 3$ is not common in given polynomials, so we ignore it and conisder only $\displaystyle x$
$\begin{array}{ r l }
& \ \underline{\ \ \ \ \ \ \ \ x-2\ \ \ \ \ \ \ }\\
x & \bigr) \ x^{2} -2x\\
& \underline{\underset{\ \ -}{} x^{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\
& \ \ \ \ \ \ \ -2x\\
& \underline{\ \ \ \ \ \ \ -2x\ \ \ \ \ \ \ \ }\\
& \ \ \ \ \ \ \ \ \ \ \ \ 0
\end{array}$
$\begin{array}{ r l }
& \ \underline{\ \ \ \ \ \ \ \ x+2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\
x^{2} -2x & \bigr) \ x^{3} -2x\\
& \underline{\underset{\ \ -}{} x^{3} \ \ \ \ \ \ \ \ \ \ \underset{+}{-} \ 2x^{2} \ }\\
& \ \ \ \ \ \ \ \ \ \ 2x^{2} \ -2x\\
& \underline{\ \ \ \ \ \ \ \underset{-}{+}2x^{2} \ \underset{+}{-}4x \ }\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x
\end{array}$
$\displaystyle 2$ is not common in given polynomials, so we ignore it and conisder only $\displaystyle x$
$\begin{array}{ r l }
& \ \underline{\ \ \ \ \ \ \ \ x-2\ \ \ \ \ \ \ }\\
x & \bigr) \ x^{2} -2x\\
& \underline{\underset{\ \ -}{} x^{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\
& \ \ \ \ \ \ \ -2x\\
& \underline{\ \ \ \ \ \ \ -2x\ \ \ \ \ \ \ \ }\\
& \ \ \ \ \ \ \ \ \ \ \ \ 0
\end{array}$
$\displaystyle \text{ HCF} =x$