Solve r^3-10r^2+32r-32=0

$$r ^ { 3 } - 10 r ^ { 2 } + 32 r - 32 = 0$$

Solve for r

$$r=4$$

$$r=2$$

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 -32 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±32,±16,±8,±4,±2,±1$$

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.

$$r=2$$

By Factor theorem, r-k is a factor of the polynomial for each root k. Divide r^{3}-10r^{2}+32r-32 by r-2 to get r^{2}-8r+16. Solve the equation where the result equals to 0.

$$r^{2}-8r+16=0$$

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, -8 for b, and 16 for c in the quadratic formula.

$$r=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$r=\frac{8±0}{2}$$

Solutions are the same.

$$r=4$$

List all found solutions.

$$r=2 r=4$$