Solve x^6-9x^4-16x^2+144=0

$${ x }^{ 6 } -9 { x }^{ 4 } -16 { x }^{ 2 } +144 = 0$$

Solve for x

$$x=-3$$

$$x=3$$

$$x=2$$

$$x=-2$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 144 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±144,±72,±48,±36,±24,±18,±16,±12,±9,±8,±6,±4,±3,±2,±1$$

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^{6}-9x^{4}-16x^{2}+144 by x-2 to get x^{5}+2x^{4}-5x^{3}-10x^{2}-36x-72. Solve the equation where the result equals to 0.

$$x^{5}+2x^{4}-5x^{3}-10x^{2}-36x-72=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -72 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±72,±36,±24,±18,±12,±9,±8,±6,±4,±3,±2,±1$$

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^{5}+2x^{4}-5x^{3}-10x^{2}-36x-72 by x+2 to get x^{4}-5x^{2}-36. Solve the equation where the result equals to 0.

$$x^{4}-5x^{2}-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$x=3$$

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

$$x^{3}+3x^{2}+4x+12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}+4x+12 by x+3 to get x^{2}+4. Solve the equation where the result equals to 0.

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

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 4 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-16}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=2 x=-2 x=3 x=-3$$

Solve for x (complex solution)

$$x\in 2i,-2i,-3,3,-2,2$$