Question

find the value of a, if x cube -ax square + 6x-a is exactly divisible by (x+3).

Answer 1

If x^3 - a x^2 + 6x - a is exactly divisible by (x+3) then
 x^3 - a x^2 + 6x - a =0
It is divisible by (x + 3) Then
x + 3 =0
x = -3
p(x) = x^3 - a x^2 + 6x - a = 0
p(-3)= (-3)^3 -a (-3)^2 + 6 (-3) - a =0
          = -27 - 9a - 18 - a = 0
          = -27 -18 -9a - a = 0
          = -45 - 10a = 0
           = -10a = 45
           = -a = 45 / 10
           = -a = 4.5
           = a = -4.5