Solve x^4-6x^3-8x^2+96x-128

$$x ^ { 4 } - 6 x ^ { 3 } - 8 x ^ { 2 } + 96 x - 128$$

Factor

$$\left(x-2\right)\left(x+4\right)\left(x-4\right)^{2}$$

Solution Steps

To factor the expression, solve the equation where it equals to 0.

$$x^{4}-6x^{3}-8x^{2}+96x-128=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -128 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±128,±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-6x^{3}-8x^{2}+96x-128 by x-2 to get x^{3}-4x^{2}-16x+64. To factor the result, solve the equation where it equals to 0.

$$x^{3}-4x^{2}-16x+64=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 64 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-4x^{2}-16x+64 by x-4 to get x^{2}-16. To factor the result, solve the equation where it equals to 0.

$$x^{2}-16=0$$

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 -16 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-16\right)}}{2}$$

Do the calculations.

$$x=\frac{0±8}{2}$$

Solve the equation x^{2}-16=0 when ± is plus and when ± is minus.

$$x=-4 x=4$$

Rewrite the factored expression using the obtained roots.

$$\left(x-2\right)\left(x+4\right)\left(x-4\right)^{2}$$

Evaluate

$$\left(x-2\right)\left(x+4\right)\left(x-4\right)^{2}$$