Solve left(x+2right)^3-23=41

$${ \left(x+2 \right) }^{ 3 } -23 = 41$$

Solve for x (complex solution)

$$x=2$$

$$x=-4+2\sqrt{3}i\approx -4+3.464101615i$$

$$x=-2\sqrt{3}i-4\approx -4-3.464101615i$$

Solution Steps

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

$$x^{3}+6x^{2}+12x+8-23=41$$

Subtract 23 from 8 to get -15.

$$x^{3}+6x^{2}+12x-15=41$$

Subtract 41 from both sides.

$$x^{3}+6x^{2}+12x-15-41=0$$

Subtract 41 from -15 to get -56.

$$x^{3}+6x^{2}+12x-56=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -56 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±56,±28,±14,±8,±7,±4,±2,±1$$

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}+6x^{2}+12x-56 by x-2 to get x^{2}+8x+28. Solve the equation where the result equals to 0.

$$x^{2}+8x+28=0$$

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, 8 for b, and 28 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 1\times 28}}{2}$$

Do the calculations.

$$x=\frac{-8±\sqrt{-48}}{2}$$

Solve the equation x^{2}+8x+28=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=2$$

Solution Steps

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

$$x^{3}+6x^{2}+12x+8-23=41$$

Subtract 23 from 8 to get -15.

$$x^{3}+6x^{2}+12x-15=41$$

Subtract 41 from both sides.

$$x^{3}+6x^{2}+12x-15-41=0$$

Subtract 41 from -15 to get -56.

$$x^{3}+6x^{2}+12x-56=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -56 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±56,±28,±14,±8,±7,±4,±2,±1$$

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}+6x^{2}+12x-56 by x-2 to get x^{2}+8x+28. Solve the equation where the result equals to 0.

$$x^{2}+8x+28=0$$

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, 8 for b, and 28 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 1\times 28}}{2}$$

Do the calculations.

$$x=\frac{-8±\sqrt{-48}}{2}$$

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.

$$x=2$$