Answer:
k=1
Step-by-step explanation:
According to the remainder Theorem, if we have a polynomial then if it divided by (x-a), the remained will be equal to
[tex]f(a)[/tex]
So we know that the factor is
[tex](x - k)[/tex]
So plug in k into the polynomial
[tex]f(k) = k {}^{3} + k(k) {}^{2} - 2(k) + 1 = k[/tex]
[tex]k {}^{3} + k {}^{3} - 2k + 1 = k[/tex]
[tex]2k {}^{3} - 3k + 1 = 0[/tex]
Solve for k.
Using Rational Roots Theorem, one possible root is 1 so we have
[tex]2k {}^{2} + 2k - 1 = 0[/tex]
Since that is irreducible over reals, 1 is our only factor so k=1.
