Solve u^3-8u^2+11u+20=0

$$u ^ { 3 } - 8 u ^ { 2 } + 11 u + 20 = 0$$

Solve for u

$$u=4$$

$$u=5$$

$$u=-1$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$u=-1$$

By Factor theorem, u-k is a factor of the polynomial for each root k. Divide u^{3}-8u^{2}+11u+20 by u+1 to get u^{2}-9u+20. Solve the equation where the result equals to 0.

$$u^{2}-9u+20=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -9 for b, and 20 for c in the quadratic formula.

$$u=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 1\times 20}}{2}$$

Do the calculations.

$$u=\frac{9±1}{2}$$

Solve the equation u^{2}-9u+20=0 when ± is plus and when ± is minus.

$$u=4 u=5$$

List all found solutions.

$$u=-1 u=4 u=5$$