Solve (x+6)(x+3)(x-1)(x-2)=12x^2

$$(x+6)(x+3)(x-1)(x-2)=12 { x }^{ 2 }$$

Solve for x

$$x=\frac{\sqrt{73}-7}{2}\approx 0.772001873$$

$$x=\frac{-\sqrt{73}-7}{2}\approx -7.772001873$$

$$x=3$$

$$x=-2$$

Solution Steps

Use the distributive property to multiply x+6 by x+3 and combine like terms.

$$\left(x^{2}+9x+18\right)\left(x-1\right)\left(x-2\right)=12x^{2}$$

Use the distributive property to multiply x^{2}+9x+18 by x-1 and combine like terms.

$$\left(x^{3}+8x^{2}+9x-18\right)\left(x-2\right)=12x^{2}$$

Use the distributive property to multiply x^{3}+8x^{2}+9x-18 by x-2 and combine like terms.

$$x^{4}+6x^{3}-7x^{2}-36x+36=12x^{2}$$

Subtract 12x^{2} from both sides.

$$x^{4}+6x^{3}-7x^{2}-36x+36-12x^{2}=0$$

Combine -7x^{2} and -12x^{2} to get -19x^{2}.

$$x^{4}+6x^{3}-19x^{2}-36x+36=0$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+6x^{3}-19x^{2}-36x+36 by x+2 to get x^{3}+4x^{2}-27x+18. Solve the equation where the result equals to 0.

$$x^{3}+4x^{2}-27x+18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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}+4x^{2}-27x+18 by x-3 to get x^{2}+7x-6. Solve the equation where the result equals to 0.

$$x^{2}+7x-6=0$$

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, 7 for b, and -6 for c in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 1\left(-6\right)}}{2}$$

Do the calculations.

$$x=\frac{-7±\sqrt{73}}{2}$$

Solve the equation x^{2}+7x-6=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{73}-7}{2} x=\frac{\sqrt{73}-7}{2}$$

List all found solutions.

$$x=-2 x=3 x=\frac{-\sqrt{73}-7}{2} x=\frac{\sqrt{73}-7}{2}$$