Solve 140b^3-132b^2+16b

$$140 b ^ { 3 } - 132 b ^ { 2 } + 16 b$$

Factor

$$4b\left(5b-4\right)\left(7b-1\right)$$

Solution Steps

Factor out 4.

$$4\left(35b^{3}-33b^{2}+4b\right)$$

Consider 35b^{3}-33b^{2}+4b. Factor out b.

$$b\left(35b^{2}-33b+4\right)$$

Consider 35b^{2}-33b+4. Factor the expression by grouping. First, the expression needs to be rewritten as 35b^{2}+pb+qb+4. To find p and q, set up a system to be solved.

$$p+q=-33 pq=35\times 4=140$$

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 140.

$$-1,-140 -2,-70 -4,-35 -5,-28 -7,-20 -10,-14$$

Calculate the sum for each pair.

$$-1-140=-141 -2-70=-72 -4-35=-39 -5-28=-33 -7-20=-27 -10-14=-24$$

The solution is the pair that gives sum -33.

$$p=-28 q=-5$$

Rewrite 35b^{2}-33b+4 as \left(35b^{2}-28b\right)+\left(-5b+4\right).

$$\left(35b^{2}-28b\right)+\left(-5b+4\right)$$

Factor out 7b in the first and -1 in the second group.

$$7b\left(5b-4\right)-\left(5b-4\right)$$

Factor out common term 5b-4 by using distributive property.

$$\left(5b-4\right)\left(7b-1\right)$$

Rewrite the complete factored expression.

$$4b\left(5b-4\right)\left(7b-1\right)$$

Evaluate

$$4b\left(5b-4\right)\left(7b-1\right)$$