Solve x^3-2x^2+5x-6=-4^2+14x+12

$${ x }^{ 3 } -2 { x }^{ 2 } +5x-6=- { 4 }^{ 2 } +14x+12$$

Solve for x

$$x=\sqrt{5}+2\approx 4.236067977$$

$$x=2-\sqrt{5}\approx -0.236067977$$

$$x=-2$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$x^{3}-2x^{2}+5x-6=-16+14x+12$$

Add -16 and 12 to get -4.

$$x^{3}-2x^{2}+5x-6=-4+14x$$

Subtract -4 from both sides.

$$x^{3}-2x^{2}+5x-6-\left(-4\right)=14x$$

The opposite of -4 is 4.

$$x^{3}-2x^{2}+5x-6+4=14x$$

Subtract 14x from both sides.

$$x^{3}-2x^{2}+5x-6+4-14x=0$$

Add -6 and 4 to get -2.

$$x^{3}-2x^{2}+5x-2-14x=0$$

Combine 5x and -14x to get -9x.

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

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}-2x^{2}-9x-2 by x+2 to get x^{2}-4x-1. Solve the equation where the result equals to 0.

$$x^{2}-4x-1=0$$

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, -4 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±2\sqrt{5}}{2}$$

Solve the equation x^{2}-4x-1=0 when ± is plus and when ± is minus.

$$x=2-\sqrt{5} x=\sqrt{5}+2$$

List all found solutions.

$$x=-2 x=2-\sqrt{5} x=\sqrt{5}+2$$