Solve (x^2+3x)^2-2(x^2+3x)=8

$$( x ^ { 2 } + 3 x ) ^ { 2 } - 2 ( x ^ { 2 } + 3 x ) = 8$$

Solve for x

$$x=-4$$

$$x=1$$

$$x=-1$$

$$x=-2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+3x\right)^{2}.

$$\left(x^{2}\right)^{2}+6x^{2}x+9x^{2}-2\left(x^{2}+3x\right)=8$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}+6x^{2}x+9x^{2}-2\left(x^{2}+3x\right)=8$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$x^{4}+6x^{3}+9x^{2}-2\left(x^{2}+3x\right)=8$$

Use the distributive property to multiply -2 by x^{2}+3x.

$$x^{4}+6x^{3}+9x^{2}-2x^{2}-6x=8$$

Combine 9x^{2} and -2x^{2} to get 7x^{2}.

$$x^{4}+6x^{3}+7x^{2}-6x=8$$

Subtract 8 from both sides.

$$x^{4}+6x^{3}+7x^{2}-6x-8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+6x^{3}+7x^{2}-6x-8 by x-1 to get x^{3}+7x^{2}+14x+8. Solve the equation where the result equals to 0.

$$x^{3}+7x^{2}+14x+8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+7x^{2}+14x+8 by x+1 to get x^{2}+6x+8. Solve the equation where the result equals to 0.

$$x^{2}+6x+8=0$$

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, 6 for b, and 8 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$x=\frac{-6±2}{2}$$

Solve the equation x^{2}+6x+8=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=1 x=-1 x=-4 x=-2$$