Solve x^3-6x+5=-3(x-1)

$$x ^ { 3 } - 6 x + 5 = - 3 ( x - 1 )$$

Solve for x

$$x=1$$

$$x=-2$$

Solution Steps

Use the distributive property to multiply -3 by x-1.

$$x^{3}-6x+5=-3x+3$$

Add 3x to both sides.

$$x^{3}-6x+5+3x=3$$

Combine -6x and 3x to get -3x.

$$x^{3}-3x+5=3$$

Subtract 3 from both sides.

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

Subtract 3 from 5 to get 2.

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

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.

$$x=1$$

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

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

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, 1 for b, and -2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±3}{2}$$

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

$$x=-2 x=1$$

List all found solutions.

$$x=1 x=-2$$