Solve (x^2-3x)^2-2(x^2-3x)-8=0

$$( x ^ { 2 } - 3 x ) ^ { 2 } - 2 ( x ^ { 2 } - 3 x ) - 8 = 0$$

Solve for x

$$x=-1$$

$$x=4$$

$$x=2$$

$$x=1$$

Solution Steps

Simplify.

$$x^{4}-6x^{3}+7x^{2}+6x-8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

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^{4}-6x^{3}+7x^{2}+6x-8 by x-1 to get x^{3}-5x^{2}+2x+8. Solve the equation where the result equals to 0.

$$x^{3}-5x^{2}+2x+8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

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}-5x^{2}+2x+8 by x+1 to get x^{2}-6x+8. Solve the equation where the result equals to 0.

$$x^{2}-6x+8=0$$

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, -6 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$x=\frac{6±2}{2}$$

Solve the equation x^{2}-6x+8=0 when ± is plus and when ± is minus.

$$x=2 x=4$$

List all found solutions.

$$x=1 x=-1 x=2 x=4$$