Solve -6ab^2+105ab-375a

$$- 6 a b ^ { 2 } + 105 a b - 375 a$$

Factor

$$3a\left(5-b\right)\left(2b-25\right)$$

Solution Steps

Factor out 3.

$$3\left(-2ab^{2}+35ab-125a\right)$$

Consider -2ab^{2}+35ab-125a. Factor out a.

$$a\left(-2b^{2}+35b-125\right)$$

Consider -2b^{2}+35b-125. Factor the expression by grouping. First, the expression needs to be rewritten as -2b^{2}+pb+qb-125. To find p and q, set up a system to be solved.

$$p+q=35 pq=-2\left(-125\right)=250$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 250.

$$1,250 2,125 5,50 10,25$$

Calculate the sum for each pair.

$$1+250=251 2+125=127 5+50=55 10+25=35$$

The solution is the pair that gives sum 35.

$$p=25 q=10$$

Rewrite -2b^{2}+35b-125 as \left(-2b^{2}+25b\right)+\left(10b-125\right).

$$\left(-2b^{2}+25b\right)+\left(10b-125\right)$$

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

$$-b\left(2b-25\right)+5\left(2b-25\right)$$

Factor out common term 2b-25 by using distributive property.

$$\left(2b-25\right)\left(-b+5\right)$$

Rewrite the complete factored expression.

$$3a\left(2b-25\right)\left(-b+5\right)$$

Evaluate

$$3a\left(5-b\right)\left(2b-25\right)$$