Solve (x^2+3x+2)(x^2-6x+8)/x^2-4=3x-12

$$\frac { ( x ^ { 2 } + 3 x + 2 ) ( x ^ { 2 } - 6 x + 8 ) } { x ^ { 2 } - 4 } = 3 x - 12$$

Solve for x

$$x=4$$

Solution Steps

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right).

$$\left(x^{2}+3x+2\right)\left(x^{2}-6x+8\right)=3x\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(x+2\right)\left(-12\right)$$

Use the distributive property to multiply x^{2}+3x+2 by x^{2}-6x+8 and combine like terms.

$$x^{4}-3x^{3}-8x^{2}+12x+16=3x\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(x+2\right)\left(-12\right)$$

Use the distributive property to multiply 3x by x-2.

$$x^{4}-3x^{3}-8x^{2}+12x+16=\left(3x^{2}-6x\right)\left(x+2\right)+\left(x-2\right)\left(x+2\right)\left(-12\right)$$

Use the distributive property to multiply 3x^{2}-6x by x+2 and combine like terms.

$$x^{4}-3x^{3}-8x^{2}+12x+16=3x^{3}-12x+\left(x-2\right)\left(x+2\right)\left(-12\right)$$

Use the distributive property to multiply x-2 by x+2 and combine like terms.

$$x^{4}-3x^{3}-8x^{2}+12x+16=3x^{3}-12x+\left(x^{2}-4\right)\left(-12\right)$$

Use the distributive property to multiply x^{2}-4 by -12.

$$x^{4}-3x^{3}-8x^{2}+12x+16=3x^{3}-12x-12x^{2}+48$$

Subtract 3x^{3} from both sides.

$$x^{4}-3x^{3}-8x^{2}+12x+16-3x^{3}=-12x-12x^{2}+48$$

Combine -3x^{3} and -3x^{3} to get -6x^{3}.

$$x^{4}-6x^{3}-8x^{2}+12x+16=-12x-12x^{2}+48$$

Add 12x to both sides.

$$x^{4}-6x^{3}-8x^{2}+12x+16+12x=-12x^{2}+48$$

Combine 12x and 12x to get 24x.

$$x^{4}-6x^{3}-8x^{2}+24x+16=-12x^{2}+48$$

Add 12x^{2} to both sides.

$$x^{4}-6x^{3}-8x^{2}+24x+16+12x^{2}=48$$

Combine -8x^{2} and 12x^{2} to get 4x^{2}.

$$x^{4}-6x^{3}+4x^{2}+24x+16=48$$

Subtract 48 from both sides.

$$x^{4}-6x^{3}+4x^{2}+24x+16-48=0$$

Subtract 48 from 16 to get -32.

$$x^{4}-6x^{3}+4x^{2}+24x-32=0$$

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}-6x^{3}+4x^{2}+24x-32 by x-2 to get x^{3}-4x^{2}-4x+16. Solve the equation where the result equals to 0.

$$x^{3}-4x^{2}-4x+16=0$$

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}-4x^{2}-4x+16 by x-2 to get x^{2}-2x-8. Solve the equation where the result equals to 0.

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

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, -2 for b, and -8 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±6}{2}$$

Solve the equation x^{2}-2x-8=0 when ± is plus and when ± is minus.

$$x=-2 x=4$$

Remove the values that the variable cannot be equal to.

$$x=4$$

List all found solutions.

$$x=2 x=-2 x=4$$

Variable x cannot be equal to any of the values 2,-2.

$$x=4$$