Solve x^4-6x^3+13x^2-24x+36=0

$${ x }^{ 4 } -6 { x }^{ 3 } +13 { x }^{ 2 } -24x+36=0$$

Solve for x

$$x=3$$

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 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}-6x^{3}+13x^{2}-24x+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.

$$x=3$$

Solve for x (complex solution)

$$x=2i$$

$$x=-2i$$

$$x=3$$