Solve 3x^3+2x^2-32=0

$$3 { x }^{ 3 } +2 { x }^{ 2 } -32 = 0$$

Solve for x (complex solution)

$$x=\frac{-4\sqrt{2}i-4}{3}\approx -1.333333333-1.885618083i$$

$$x=2$$

$$x=\frac{-4+4\sqrt{2}i}{3}\approx -1.333333333+1.885618083i$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -32 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{32}{3},±32,±\frac{16}{3},±16,±\frac{8}{3},±8,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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 3x^{3}+2x^{2}-32 by x-2 to get 3x^{2}+8x+16. Solve the equation where the result equals to 0.

$$3x^{2}+8x+16=0$$

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 3 for a, 8 for b, and 16 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\times 16}}{2\times 3}$$

Do the calculations.

$$x=\frac{-8±\sqrt{-128}}{6}$$

Solve the equation 3x^{2}+8x+16=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=2$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -32 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{32}{3},±32,±\frac{16}{3},±16,±\frac{8}{3},±8,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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 3x^{3}+2x^{2}-32 by x-2 to get 3x^{2}+8x+16. Solve the equation where the result equals to 0.

$$3x^{2}+8x+16=0$$

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 3 for a, 8 for b, and 16 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\times 16}}{2\times 3}$$

Do the calculations.

$$x=\frac{-8±\sqrt{-128}}{6}$$

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.

$$x=2$$