Solve a^3-2a^2-a+7=5

$$a ^ { 3 } - 2 a ^ { 2 } - a + 7 = 5$$

Solve for a

$$a=1$$

$$a=2$$

$$a=-1$$

Solution Steps

Subtract 5 from both sides.

$$a^{3}-2a^{2}-a+7-5=0$$

Subtract 5 from 7 to get 2.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

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.

$$a=1$$

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

$$a^{2}-a-2=0$$

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 -2 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{1±3}{2}$$

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

$$a=-1 a=2$$

List all found solutions.

$$a=1 a=-1 a=2$$