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