Solve x^3-6x^2+9x-4=0

$${ x }^{ 3 } -6 { x }^{ 2 } +9x-4 = 0$$

Solve for x

$$x=1$$

$$x=4$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-6x^{2}+9x-4 by x-1 to get x^{2}-5x+4. Solve the equation where the result equals to 0.

$$x^{2}-5x+4=0$$

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, -5 for b, and 4 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{5±3}{2}$$

Solve the equation x^{2}-5x+4=0 when ± is plus and when ± is minus.

$$x=1 x=4$$

List all found solutions.

$$x=1 x=4$$