Solve x^3+3x^2-9x+1=3

$$x ^ { 3 } + 3 x ^ { 2 } - 9 x + 1 = 3$$

Solve for x

$$x=\frac{-\sqrt{21}-5}{2}\approx -4.791287847$$

$$x=2$$

$$x=\frac{\sqrt{21}-5}{2}\approx -0.208712153$$

Solution Steps

Subtract 3 from both sides.

$$x^{3}+3x^{2}-9x+1-3=0$$

Subtract 3 from 1 to get -2.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}-9x-2 by x-2 to get x^{2}+5x+1. Solve the equation where the result equals to 0.

$$x^{2}+5x+1=0$$

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, 5 for b, and 1 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{21}}{2}$$

Solve the equation x^{2}+5x+1=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{21}-5}{2} x=\frac{\sqrt{21}-5}{2}$$

List all found solutions.

$$x=2 x=\frac{-\sqrt{21}-5}{2} x=\frac{\sqrt{21}-5}{2}$$