Question

Find the roots of the polynomial equation. x^3-x^2+x+39

Answer 1

Use the rational zero theorem to find the possible roots. So the coefficient of x^3 is 1 and the last term is 39.

So the factors of 1 are +1, -1 and the factors of 39 are -3,3,-13,13, 39, -39, 1,-1

Since we are dividing the factors of 39 by the factors of 1, we just need to test the factors of 39 by substituting the values for x in the equation.

When you do that, you should find that -3 is the only real zero of the equation.

So divide the equation by x+3\[\frac{ (x ^{3}-x ^{2}+x+39) }{ (x+3) }\]\[x ^{2}-4x+13\]

Solve using the quadratic equation.

\[\frac{ (4\pm \sqrt{-4^{2}-4*1*13}) }{ 2}\]\[\frac{ (4\pm \sqrt{-36} }{ 2 }\]\[\frac{ (4\pm6i }{ 2 }\]\[2\pm3i\]

So the roots are (x+3)(x-(2-3i))(x-(2+3i))