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

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

Solve for u

$$u=1$$

Solution Steps

Swap sides so that all variable terms are on the left hand side.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -3 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±3,±1$$

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}-2u^{2}+4u-3 by u-1 to get u^{2}-u+3. Solve the equation where the result equals to 0.

$$u^{2}-u+3=0$$

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, -1 for b, and 3 for c in the quadratic formula.

$$u=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$u=\frac{1±\sqrt{-11}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$u\in \emptyset$$

List all found solutions.

$$u=1$$