Solve 7b^3-21b^2-1666b

$$7 b ^ { 3 } - 21 b ^ { 2 } - 1666 b$$

Factor

$$7b\left(b-17\right)\left(b+14\right)$$

Solution Steps

Factor out 7.

$$7\left(b^{3}-3b^{2}-238b\right)$$

Consider b^{3}-3b^{2}-238b. Factor out b.

$$b\left(b^{2}-3b-238\right)$$

Consider b^{2}-3b-238. Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb-238. To find p and q, set up a system to be solved.

$$p+q=-3 pq=1\left(-238\right)=-238$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -238.

$$1,-238 2,-119 7,-34 14,-17$$

Calculate the sum for each pair.

$$1-238=-237 2-119=-117 7-34=-27 14-17=-3$$

The solution is the pair that gives sum -3.

$$p=-17 q=14$$

Rewrite b^{2}-3b-238 as \left(b^{2}-17b\right)+\left(14b-238\right).

$$\left(b^{2}-17b\right)+\left(14b-238\right)$$

Factor out b in the first and 14 in the second group.

$$b\left(b-17\right)+14\left(b-17\right)$$

Factor out common term b-17 by using distributive property.

$$\left(b-17\right)\left(b+14\right)$$

Rewrite the complete factored expression.

$$7b\left(b-17\right)\left(b+14\right)$$

Evaluate

$$7b\left(b-17\right)\left(b+14\right)$$