Solve 10x^2+x^3-6=3

$$10 x ^ { 2 } + x ^ { 3 } - 6 = 3$$

Solve for x

$$x=\frac{-3\sqrt{13}-9}{2}\approx -9.908326913$$

$$x=-1$$

$$x=\frac{3\sqrt{13}-9}{2}\approx 0.908326913$$

Solution Steps

Subtract 3 from both sides.

$$10x^{2}+x^{3}-6-3=0$$

Subtract 3 from -6 to get -9.

$$10x^{2}+x^{3}-9=0$$

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

$$x^{3}+10x^{2}-9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+10x^{2}-9 by x+1 to get x^{2}+9x-9. Solve the equation where the result equals to 0.

$$x^{2}+9x-9=0$$

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, 9 for b, and -9 for c in the quadratic formula.

$$x=\frac{-9±\sqrt{9^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$x=\frac{-9±3\sqrt{13}}{2}$$

Solve the equation x^{2}+9x-9=0 when ± is plus and when ± is minus.

$$x=\frac{-3\sqrt{13}-9}{2} x=\frac{3\sqrt{13}-9}{2}$$

List all found solutions.

$$x=-1 x=\frac{-3\sqrt{13}-9}{2} x=\frac{3\sqrt{13}-9}{2}$$