Set equal to .

Factor using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form where is a factor of the constant and is a factor of the leading coefficient.

Find every combination of . These are the possible roots of the polynomial function.

Substitute and simplify the expression. In this case, the expression is equal to so is a root of the polynomial.

Substitute into the polynomial.

Raise to the power of .

Raise to the power of .

Multiply by .

Subtract from .

Multiply by .

Subtract from .

Add and .

Since is a known root, divide the polynomial by to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

Divide by .

Write as a set of factors.

If any individual factor on the left side of the equation is equal to , the entire expression will be equal to .

Set the first factor equal to and solve.

Set the first factor equal to .

Subtract from both sides of the equation.

Set the next factor equal to and solve.

Set the next factor equal to .

Use the quadratic formula to find the solutions.

Substitute the values , , and into the quadratic formula and solve for .

Simplify.

Simplify the numerator.

Raise to the power of .

Multiply by .

Multiply by .

Subtract from .

Rewrite as .

Rewrite as .

Rewrite as .

Rewrite as .

Pull terms out from under the radical, assuming positive real numbers.

Move to the left of .

Multiply by .

Simplify .

The final answer is the combination of both solutions.

The final solution is all the values that make true.

Find the Roots (Zeros) x^3-2x^2+10x+136