Solve x^6-5x^4+16x^3+36x^2-16x-32

$${ x }^{ 6 } -5 { x }^{ 4 } +16 { x }^{ 3 } +36 { x }^{ 2 } -16x-32$$

Factor

$$\left(x-1\right)\left(x+1\right)\left(x+2\right)^{2}\left(x^{2}-4x+8\right)$$

Solution Steps

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

$$x^{6}-5x^{4}+16x^{3}+36x^{2}-16x-32=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -32 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±32,±16,±8,±4,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{6}-5x^{4}+16x^{3}+36x^{2}-16x-32 by x-1 to get x^{5}+x^{4}-4x^{3}+12x^{2}+48x+32. To factor the result, solve the equation where it equals to 0.

$$x^{5}+x^{4}-4x^{3}+12x^{2}+48x+32=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 32 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±32,±16,±8,±4,±2,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{5}+x^{4}-4x^{3}+12x^{2}+48x+32 by x+1 to get x^{4}-4x^{2}+16x+32. To factor the result, solve the equation where it equals to 0.

$$x^{4}-4x^{2}+16x+32=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 32 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±32,±16,±8,±4,±2,±1$$

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}-4x^{2}+16x+32 by x+2 to get x^{3}-2x^{2}+16. To factor the result, solve the equation where it equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 16 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±16,±8,±4,±2,±1$$

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^{3}-2x^{2}+16 by x+2 to get x^{2}-4x+8. To factor the result, solve the equation where it equals to 0.

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

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, -4 for b, and 8 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±\sqrt{-16}}{2}$$

Polynomial x^{2}-4x+8 is not factored since it does not have any rational roots.

$$x^{2}-4x+8$$

Rewrite the factored expression using the obtained roots.

$$\left(x-1\right)\left(x+1\right)\left(x+2\right)^{2}\left(x^{2}-4x+8\right)$$

Evaluate

$$\left(x^{2}-1\right)\left(x+2\right)^{2}\left(x^{2}-4x+8\right)$$