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3 years ago

2. If k is a constant, what is the value of k such that the polynomial k2x3 - 6kx+9 is divisible by x - 1?

Mathematics

1 answer:

marysya [2.9K]3 years ago

4 0

Answer: The value of k =3.

Step-by-step explanation:

According to factor theorem, if g(x) =x-a is a factor of p(x) then p(a)=0.

Given: $x-1$ is a factor of $k^2x^3-6kx+9$ .

then

$k^2(1)^3-6k(1)+9=0\\\\\Rightarrow\ k^2-6k+9=0\\\\\Rightarrow\ k^2-3k-3k+9=0\\\\\Rightarrow\ k(k-3)-3(k-3)=0\\\\\Rightarrow\ (k-3)(k-3)=0\\\\\Rightarrow\ k=3$

Hence, the value of k =3.

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