Solve r^3-4r^2+r+6=0

$$r ^ { 3 } - 4 r ^ { 2 } + r + 6 = 0$$

Solve for r

$$r=2$$

$$r=3$$

$$r=-1$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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.

$$r=-1$$

By Factor theorem, r-k is a factor of the polynomial for each root k. Divide r^{3}-4r^{2}+r+6 by r+1 to get r^{2}-5r+6. Solve the equation where the result equals to 0.

$$r^{2}-5r+6=0$$

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 6 for c in the quadratic formula.

$$r=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$r=\frac{5±1}{2}$$

Solve the equation r^{2}-5r+6=0 when ± is plus and when ± is minus.

$$r=2 r=3$$

List all found solutions.

$$r=-1 r=2 r=3$$