Solve x^3-x^2-x-1=-2

$$x ^ { 3 } - x ^ { 2 } - x - 1 = - 2$$

Solve for x

$$x=1$$

$$x=-1$$

Solution Steps

Add 2 to both sides.

$$x^{3}-x^{2}-x-1+2=0$$

Add -1 and 2 to get 1.

$$x^{3}-x^{2}-x+1=0$$

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

$$±1$$

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

$$x^{2}-1=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-1\right)}}{2}$$

Do the calculations.

$$x=\frac{0±2}{2}$$

Solve the equation x^{2}-1=0 when ± is plus and when ± is minus.

$$x=-1 x=1$$

List all found solutions.

$$x=1 x=-1$$