Question

if x3-6x2+6x +k is completely divisible by x-3 then find the value of k

Answer 1

Heya

Let p(x) = x³-6x²+6x+k

And g(x) = x-3

It is given that p(x) is completely divisible by g(x)

=> g(x) is a factor of p(x)

Using factor theorem which states as, if p(a) is a factor of g(x) then g(a)=0

g(x) = x-3

x= 3

p(3) = 0

(3)³-6(3)²+6(3)+k = 0

27-54+18+k = 0

-9+k =0

kis 9

Answer 2

Heya !!!




Given that ( X -3) is a factor of the given polynomial.


So,


( X-3) = 0


X = 3



P(X) = X³ -6X² + 6X + K



P(3) = (3)³ - 6 × (3)² + 6 × 3 + K




=> 27 - 6 × 9 + 18 + K




=> 27 - 54 + 18 + K = 0




=> 45 -54 + K = 0




=> K -9 = 0



=> K = 9




HOPE IT WILL HELP YOU...... :-)