Solve (x^2+3x-2)(x^2+3x+4)=16

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

Solve for x (complex solution)

$$x=1$$

$$x=-4$$

$$x=\frac{-3+\sqrt{15}i}{2}\approx -1.5+1.936491673i$$

$$x=\frac{-\sqrt{15}i-3}{2}\approx -1.5-1.936491673i$$

Solution Steps

Use the distributive property to multiply x^{2}+3x-2 by x^{2}+3x+4 and combine like terms.

$$x^{4}+6x^{3}+11x^{2}+6x-8=16$$

Subtract 16 from both sides.

$$x^{4}+6x^{3}+11x^{2}+6x-8-16=0$$

Subtract 16 from -8 to get -24.

$$x^{4}+6x^{3}+11x^{2}+6x-24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+6x^{3}+11x^{2}+6x-24 by x-1 to get x^{3}+7x^{2}+18x+24. Solve the equation where the result equals to 0.

$$x^{3}+7x^{2}+18x+24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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}+7x^{2}+18x+24 by x+4 to get x^{2}+3x+6. Solve the equation where the result equals to 0.

$$x^{2}+3x+6=0$$

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-15}}{2}$$

Solve the equation x^{2}+3x+6=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{15}i-3}{2} x=\frac{-3+\sqrt{15}i}{2}$$

List all found solutions.

$$x=1 x=-4 x=\frac{-\sqrt{15}i-3}{2} x=\frac{-3+\sqrt{15}i}{2}$$

Solve for x

$$x=-4$$

$$x=1$$

Solution Steps

Use the distributive property to multiply x^{2}+3x-2 by x^{2}+3x+4 and combine like terms.

$$x^{4}+6x^{3}+11x^{2}+6x-8=16$$

Subtract 16 from both sides.

$$x^{4}+6x^{3}+11x^{2}+6x-8-16=0$$

Subtract 16 from -8 to get -24.

$$x^{4}+6x^{3}+11x^{2}+6x-24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+6x^{3}+11x^{2}+6x-24 by x-1 to get x^{3}+7x^{2}+18x+24. Solve the equation where the result equals to 0.

$$x^{3}+7x^{2}+18x+24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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}+7x^{2}+18x+24 by x+4 to get x^{2}+3x+6. Solve the equation where the result equals to 0.

$$x^{2}+3x+6=0$$

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-15}}{2}$$

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.

$$x=1 x=-4$$