Solve 100r^5+100r^4-3000r^3

$$100 r ^ { 5 } + 100 r ^ { 4 } - 3000 r ^ { 3 }$$

Factor

$$100\left(r-5\right)\left(r+6\right)r^{3}$$

Solution Steps

Factor out 100.

$$100\left(r^{5}+r^{4}-30r^{3}\right)$$

Consider r^{5}+r^{4}-30r^{3}. Factor out r^{3}.

$$r^{3}\left(r^{2}+r-30\right)$$

Consider r^{2}+r-30. Factor the expression by grouping. First, the expression needs to be rewritten as r^{2}+ar+br-30. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-30\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite r^{2}+r-30 as \left(r^{2}-5r\right)+\left(6r-30\right).

$$\left(r^{2}-5r\right)+\left(6r-30\right)$$

Factor out r in the first and 6 in the second group.

$$r\left(r-5\right)+6\left(r-5\right)$$

Factor out common term r-5 by using distributive property.

$$\left(r-5\right)\left(r+6\right)$$

Rewrite the complete factored expression.

$$100r^{3}\left(r-5\right)\left(r+6\right)$$

Evaluate

$$100\left(r-5\right)\left(r+6\right)r^{3}$$