Solve left(x+1right)left(x^2+2right)+left(x+2right)left(x^2+1right)=2

$$\left( x+1 \right) \left( { x }^{ 2 } +2 \right) + \left( x+2 \right) \left( { x }^{ 2 } +1 \right) = 2$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{15}i-1}{4}\approx -0.25-0.968245837i$$

$$x=-1$$

$$x=\frac{-1+\sqrt{15}i}{4}\approx -0.25+0.968245837i$$

Solution Steps

Use the distributive property to multiply x+1 by x^{2}+2.

$$x^{3}+2x+x^{2}+2+\left(x+2\right)\left(x^{2}+1\right)=2$$

Use the distributive property to multiply x+2 by x^{2}+1.

$$x^{3}+2x+x^{2}+2+x^{3}+x+2x^{2}+2=2$$

Combine x^{3} and x^{3} to get 2x^{3}.

$$2x^{3}+2x+x^{2}+2+x+2x^{2}+2=2$$

Combine 2x and x to get 3x.

$$2x^{3}+3x+x^{2}+2+2x^{2}+2=2$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$2x^{3}+3x+3x^{2}+2+2=2$$

Add 2 and 2 to get 4.

$$2x^{3}+3x+3x^{2}+4=2$$

Subtract 2 from both sides.

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

Subtract 2 from 4 to get 2.

$$2x^{3}+3x+3x^{2}+2=0$$

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

$$2x^{3}+3x^{2}+3x+2=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 2 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±1,±2,±\frac{1}{2}$$

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 2x^{3}+3x^{2}+3x+2 by x+1 to get 2x^{2}+x+2. Solve the equation where the result equals to 0.

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

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 2 for a, 1 for b, and 2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±\sqrt{-15}}{4}$$

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

$$x=\frac{-\sqrt{15}i-1}{4} x=\frac{-1+\sqrt{15}i}{4}$$

List all found solutions.

$$x=-1 x=\frac{-\sqrt{15}i-1}{4} x=\frac{-1+\sqrt{15}i}{4}$$

Solve for x

$$x=-1$$

Solution Steps

Use the distributive property to multiply x+1 by x^{2}+2.

$$x^{3}+2x+x^{2}+2+\left(x+2\right)\left(x^{2}+1\right)=2$$

Use the distributive property to multiply x+2 by x^{2}+1.

$$x^{3}+2x+x^{2}+2+x^{3}+x+2x^{2}+2=2$$

Combine x^{3} and x^{3} to get 2x^{3}.

$$2x^{3}+2x+x^{2}+2+x+2x^{2}+2=2$$

Combine 2x and x to get 3x.

$$2x^{3}+3x+x^{2}+2+2x^{2}+2=2$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$2x^{3}+3x+3x^{2}+2+2=2$$

Add 2 and 2 to get 4.

$$2x^{3}+3x+3x^{2}+4=2$$

Subtract 2 from both sides.

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

Subtract 2 from 4 to get 2.

$$2x^{3}+3x+3x^{2}+2=0$$

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

$$2x^{3}+3x^{2}+3x+2=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 2 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±1,±2,±\frac{1}{2}$$

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 2x^{3}+3x^{2}+3x+2 by x+1 to get 2x^{2}+x+2. Solve the equation where the result equals to 0.

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

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 2 for a, 1 for b, and 2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±\sqrt{-15}}{4}$$

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.

$$x=-1$$