Solve 3x^3+4x^2=12x+16

$$3 x ^ { 3 } + 4 x ^ { 2 } = 12 x + 16$$

Solve for x

$$x=-2$$

$$x=2$$

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

Solution Steps

Subtract 12x from both sides.

$$3x^{3}+4x^{2}-12x=16$$

Subtract 16 from both sides.

$$3x^{3}+4x^{2}-12x-16=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -16 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

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

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}+4x^{2}-12x-16 by x-2 to get 3x^{2}+10x+8. Solve the equation where the result equals to 0.

$$3x^{2}+10x+8=0$$

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

$$x=\frac{-10±\sqrt{10^{2}-4\times 3\times 8}}{2\times 3}$$

Do the calculations.

$$x=\frac{-10±2}{6}$$

Solve the equation 3x^{2}+10x+8=0 when ± is plus and when ± is minus.

$$x=-2 x=-\frac{4}{3}$$

List all found solutions.

$$x=2 x=-2 x=-\frac{4}{3}$$