Question

Find the H.C.F. of $x^3-7x^2+14x-8$ and $x^3-6x^2+11x-6$ by division method.

Answer 1

Two equations are given

$x^3-7x^2+14x-8=0$ ......... $\left(1\right)$ and $x^3-6x^2+11x-6=0$ ........ $\left(2\right)$

We have used synthetic division to solve the given equations.

From part (A), we get the factor of equation $\left(1\right)$

$\therefore$ $x^3-7x^2+14x-8=(x-1)(x^2-6x+8)$

$=(x-1)(x^2-4x-2x+8)$

$=(x-1)\left(x\right(x-4)-2(x-4\left)\right)$

$=(x-1)(x-4)(x-2)$

$\therefore$ $x^3-7x^2+14x-8=(x-1)(x-2)(x-4)$ .................. $\left(3\right)$

From part (B), we get the factor of equation $\left(2\right)$

$\therefore$ $x^3-6x^2+11x-6=(x-1)(x^2-5x+6)$

$=(x-1)(x^2-3x-2x+6)$

$=(x-1)\left(x\right(x-3)-2(x-3\left)\right)$

$=(x-1)(x-2)(x-3)$

$\therefore$ $x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$ .................. $\left(4\right)$

From equation $\left(3\right)$ and $\left(4\right)$,

$(x-1)(x-2)$ is the H.C.F(highest common factor) of given equations $\left(1\right)$ and $\left(2\right)$.