Solve 30b^3+69b^2-15b

$$30 b ^ { 3 } + 69 b ^ { 2 } - 15 b$$

Factor

$$3b\left(5b-1\right)\left(2b+5\right)$$

Solution Steps

Factor out 3.

$$3\left(10b^{3}+23b^{2}-5b\right)$$

Consider 10b^{3}+23b^{2}-5b. Factor out b.

$$b\left(10b^{2}+23b-5\right)$$

Consider 10b^{2}+23b-5. Factor the expression by grouping. First, the expression needs to be rewritten as 10b^{2}+pb+qb-5. To find p and q, set up a system to be solved.

$$p+q=23 pq=10\left(-5\right)=-50$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -50.

$$-1,50 -2,25 -5,10$$

Calculate the sum for each pair.

$$-1+50=49 -2+25=23 -5+10=5$$

The solution is the pair that gives sum 23.

$$p=-2 q=25$$

Rewrite 10b^{2}+23b-5 as \left(10b^{2}-2b\right)+\left(25b-5\right).

$$\left(10b^{2}-2b\right)+\left(25b-5\right)$$

Factor out 2b in the first and 5 in the second group.

$$2b\left(5b-1\right)+5\left(5b-1\right)$$

Factor out common term 5b-1 by using distributive property.

$$\left(5b-1\right)\left(2b+5\right)$$

Rewrite the complete factored expression.

$$3b\left(5b-1\right)\left(2b+5\right)$$

Evaluate

$$3b\left(5b-1\right)\left(2b+5\right)$$