Solve r^3-10r^2-100r+1000

$$r ^ { 3 } - 10 r ^ { 2 } - 100 r + 1000$$

Factor

$$\left(r+10\right)\left(r-10\right)^{2}$$

Solution Steps

Do the grouping r^{3}-10r^{2}-100r+1000=\left(r^{3}-10r^{2}\right)+\left(-100r+1000\right), and factor out r^{2} in the first and -100 in the second group.

$$r^{2}\left(r-10\right)-100\left(r-10\right)$$

Factor out common term r-10 by using distributive property.

$$\left(r-10\right)\left(r^{2}-100\right)$$

Consider r^{2}-100. Rewrite r^{2}-100 as r^{2}-10^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(r-10\right)\left(r+10\right)$$

Rewrite the complete factored expression.

$$\left(r+10\right)\left(r-10\right)^{2}$$

Evaluate

$$\left(r+10\right)\left(r-10\right)^{2}$$