Solve a^3+a^2-a+1=2

$$a ^ { 3 } + a ^ { 2 } - a + 1 = 2$$

Solve for a

$$a=-1$$

$$a=1$$

Solution Steps

Subtract 2 from both sides.

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

Subtract 2 from 1 to get -1.

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

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.

$$a=1$$

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

$$a^{2}+2a+1=0$$

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, 2 for b, and 1 for c in the quadratic formula.

$$a=\frac{-2±\sqrt{2^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$a=\frac{-2±0}{2}$$

Solutions are the same.

$$a=-1$$

List all found solutions.

$$a=1 a=-1$$