Solve 16-36x-x^3=36x^2-3x^4

$$16 - 36 x - x ^ { 3 } = 36 x ^ { 2 } - 3 x ^ { 4 }$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=4$$

$$x=-2$$

Solution Steps

Subtract 36x^{2} from both sides.

$$16-36x-x^{3}-36x^{2}=-3x^{4}$$

Add 3x^{4} to both sides.

$$16-36x-x^{3}-36x^{2}+3x^{4}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{4}-x^{3}-36x^{2}-36x+16=0$$

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^{4}-x^{3}-36x^{2}-36x+16 by x+2 to get 3x^{3}-7x^{2}-22x+8. Solve the equation where the result equals to 0.

$$3x^{3}-7x^{2}-22x+8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 8 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

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

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}-7x^{2}-22x+8 by x+2 to get 3x^{2}-13x+4. Solve the equation where the result equals to 0.

$$3x^{2}-13x+4=0$$

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, -13 for b, and 4 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 3\times 4}}{2\times 3}$$

Do the calculations.

$$x=\frac{13±11}{6}$$

Solve the equation 3x^{2}-13x+4=0 when ± is plus and when ± is minus.

$$x=\frac{1}{3} x=4$$

List all found solutions.

$$x=-2 x=\frac{1}{3} x=4$$