Solve q^3-3q^2+4=0

$$q ^ { 3 } - 3 q ^ { 2 } + 4 = 0$$

Solve for q

$$q=2$$

$$q=-1$$

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 4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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.

$$q=-1$$

By Factor theorem, q-k is a factor of the polynomial for each root k. Divide q^{3}-3q^{2}+4 by q+1 to get q^{2}-4q+4. Solve the equation where the result equals to 0.

$$q^{2}-4q+4=0$$

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, -4 for b, and 4 for c in the quadratic formula.

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

Do the calculations.

$$q=\frac{4±0}{2}$$

Solutions are the same.

$$q=2$$

List all found solutions.

$$q=-1 q=2$$