Question

find the value of a, if x cube - ax square+ 6x -a is exactly divisible by (x+2).

Answer 1

$\huge\boxed{\red{\bold{Answer}}}$

By remainder theorm 1st find the zeros of x+2

=> x+2=0

x=(-2)

Put the value of x in equation

$\sf x{}^{3}-ax{}^{2}+6x-a=0\\ \sf(-2){}^{3}-a(-2){}^{2}+6(-2)-a=0\\ \sf {-8}-4a-12-a=0\\ \sf -20-5a=0\\ \sf (-5a)=20\\ \sf a=\frac{20}{-5}\\ \sf a=(-4)$

The value of a is (-4)

$\boxed{\red{a=(-4)}}$

Thanks !!☺️