Solve (2x+10)^3=-27

$$( 2 x + 10 ) ^ { 3 } = - 27$$

Solve for x (complex solution)

$$x = -\frac{13}{2} = -6\frac{1}{2} = -6.5$$

$$x=\frac{-17+3\sqrt{3}i}{4}\approx -4.25+1.299038106i$$

$$x=\frac{-3\sqrt{3}i-17}{4}\approx -4.25-1.299038106i$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(2x+10\right)^{3}.

$$8x^{3}+120x^{2}+600x+1000=-27$$

Add 27 to both sides.

$$8x^{3}+120x^{2}+600x+1000+27=0$$

Add 1000 and 27 to get 1027.

$$8x^{3}+120x^{2}+600x+1027=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1027 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{1027}{8},±\frac{1027}{4},±\frac{1027}{2},±1027,±\frac{79}{8},±\frac{79}{4},±\frac{79}{2},±79,±\frac{13}{8},±\frac{13}{4},±\frac{13}{2},±13,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-\frac{13}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}+120x^{2}+600x+1027 by 2\left(x+\frac{13}{2}\right)=2x+13 to get 4x^{2}+34x+79. Solve the equation where the result equals to 0.

$$4x^{2}+34x+79=0$$

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 4 for a, 34 for b, and 79 for c in the quadratic formula.

$$x=\frac{-34±\sqrt{34^{2}-4\times 4\times 79}}{2\times 4}$$

Do the calculations.

$$x=\frac{-34±\sqrt{-108}}{8}$$

Solve the equation 4x^{2}+34x+79=0 when ± is plus and when ± is minus.

$$x=\frac{-3i\sqrt{3}-17}{4} x=\frac{-17+3i\sqrt{3}}{4}$$

List all found solutions.

$$x=-\frac{13}{2} x=\frac{-3i\sqrt{3}-17}{4} x=\frac{-17+3i\sqrt{3}}{4}$$

Solve for x

$$x = -\frac{13}{2} = -6\frac{1}{2} = -6.5$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(2x+10\right)^{3}.

$$8x^{3}+120x^{2}+600x+1000=-27$$

Add 27 to both sides.

$$8x^{3}+120x^{2}+600x+1000+27=0$$

Add 1000 and 27 to get 1027.

$$8x^{3}+120x^{2}+600x+1027=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1027 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{1027}{8},±\frac{1027}{4},±\frac{1027}{2},±1027,±\frac{79}{8},±\frac{79}{4},±\frac{79}{2},±79,±\frac{13}{8},±\frac{13}{4},±\frac{13}{2},±13,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-\frac{13}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}+120x^{2}+600x+1027 by 2\left(x+\frac{13}{2}\right)=2x+13 to get 4x^{2}+34x+79. Solve the equation where the result equals to 0.

$$4x^{2}+34x+79=0$$

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 4 for a, 34 for b, and 79 for c in the quadratic formula.

$$x=\frac{-34±\sqrt{34^{2}-4\times 4\times 79}}{2\times 4}$$

Do the calculations.

$$x=\frac{-34±\sqrt{-108}}{8}$$

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.

$$x=-\frac{13}{2}$$