Solve 6ab^2-20ab+14a

$$6 a b ^ { 2 } - 20 a b + 14 a$$

Factor

$$2a\left(b-1\right)\left(3b-7\right)$$

Solution Steps

Factor out 2.

$$2\left(3ab^{2}-10ab+7a\right)$$

Consider 3ab^{2}-10ab+7a. Factor out a.

$$a\left(3b^{2}-10b+7\right)$$

Consider 3b^{2}-10b+7. Factor the expression by grouping. First, the expression needs to be rewritten as 3b^{2}+pb+qb+7. To find p and q, set up a system to be solved.

$$p+q=-10 pq=3\times 7=21$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 21.

$$-1,-21 -3,-7$$

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -10.

$$p=-7 q=-3$$

Rewrite 3b^{2}-10b+7 as \left(3b^{2}-7b\right)+\left(-3b+7\right).

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

Factor out b in the first and -1 in the second group.

$$b\left(3b-7\right)-\left(3b-7\right)$$

Factor out common term 3b-7 by using distributive property.

$$\left(3b-7\right)\left(b-1\right)$$

Rewrite the complete factored expression.

$$2a\left(3b-7\right)\left(b-1\right)$$

Evaluate

$$2a\left(b-1\right)\left(3b-7\right)$$