Question

x^3 + 7kx^2 - 4kx + 12 is completely divisible by x-3 then find k

Answer 1

First we will find thr zero of divisior

x-3=0
x=3

Now we will find the remainder of p(x) using remainder theorem

p(x)=x^3+7x^2-4kx+12
Put x=3
p(3)=3^3+7*k*3^2-4*3*k+12
p(3)=27+7*k*9-12k+12
p(3)=27+63k+12-12k
p(3)=39+51k

Now, the remainder must be 0 to make p(x) a factor of x-3

0=39+51k
-39=51k
-39/51=k
-13/17=k

The value of k is -13/17

Answer 2

Let p(x)=x^3+7kx^2-4kx+12
Let g(x)=x-3=0
x=3
As g(x) is a factor of p(x)
R=0
Therefore,
P(3)=0
p(3)=(3)^3+7k(3)^2-4k(3)+12=0
27+63k-12k+12=0
39+51k=0
51k=-39
k=-39/51