Solve (x+2)(x-3)(x+4)(x-6)+6x^2=0

$$( x + 2 ) ( x - 3 ) ( x + 4 ) ( x - 6 ) + 6 x ^ { 2 } = 0$$

Solve for x

$$x=\sqrt{13}+1\approx 4.605551275$$

$$x=1-\sqrt{13}\approx -2.605551275$$

$$x=4$$

$$x=-3$$

Solution Steps

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$\left(x-6\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+6x^{2}=0$$

Multiply and combine like terms.

$$x^{4}-3x^{3}-22x^{2}+36x+144=0$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-3x^{3}-22x^{2}+36x+144 by x+3 to get x^{3}-6x^{2}-4x+48. Solve the equation where the result equals to 0.

$$x^{3}-6x^{2}-4x+48=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 48 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±48,±24,±16,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=4$$

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

$$x^{2}-2x-12=0$$

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, -2 for b, and -12 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$x=\frac{2±2\sqrt{13}}{2}$$

Solve the equation x^{2}-2x-12=0 when ± is plus and when ± is minus.

$$x=1-\sqrt{13} x=\sqrt{13}+1$$

List all found solutions.

$$x=-3 x=4 x=1-\sqrt{13} x=\sqrt{13}+1$$