Solve x^3-10x^2+28x-24=0

$$x ^ { 3 } - 10 x ^ { 2 } + 28 x - 24 = 0$$

Solve for x

$$x=2$$

$$x=6$$

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 -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-10x^{2}+28x-24 by x-2 to get x^{2}-8x+12. Solve the equation where the result equals to 0.

$$x^{2}-8x+12=0$$

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, -8 for b, and 12 for c in the quadratic formula.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$x=\frac{8±4}{2}$$

Solve the equation x^{2}-8x+12=0 when ± is plus and when ± is minus.

$$x=2 x=6$$

List all found solutions.

$$x=2 x=6$$