Question

The factors of $(x^2-x)-8(x^2-x)+12=0$ are

• A. $-1,1,3,2$
• B. $1,1,2,3$
• C. $-1,-2,3,-2$
• D. $-1,-2,3,2$

Answer 1

Answer: (D)

$-1,-2,3,2$

${(x^2-x)}^2-8(x^2-x)+12=0$

Say, $x^2-x=y$

$y^2-8y+12=0\Rightarrow (y-2)(y-6)=0$

$\therefore y=2$ or$6$

When $y=2$

$\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0$

$\therefore x=-1$ or$2$

When $y=6$

$x^2-x-6=0\Rightarrow (x-3)(x+2)=0$

$\therefore x=-2or3$

$\therefore$ Solutions possible are $-2,-1,2,3$

Answer:-$D$