Solve 32b^2-56b-12b+21

$$32 b ^ { 2 } - 56 b - 12 b + 21$$

Factor

$$\left(4b-7\right)\left(8b-3\right)$$

Solution Steps

Multiply and combine like terms.

$$32b^{2}-68b+21$$

Factor the expression by grouping. First, the expression needs to be rewritten as 32b^{2}+pb+qb+21. To find p and q, set up a system to be solved.

$$p+q=-68 pq=32\times 21=672$$

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 672.

$$-1,-672 -2,-336 -3,-224 -4,-168 -6,-112 -7,-96 -8,-84 -12,-56 -14,-48 -16,-42 -21,-32 -24,-28$$

Calculate the sum for each pair.

$$-1-672=-673 -2-336=-338 -3-224=-227 -4-168=-172 -6-112=-118 -7-96=-103 -8-84=-92 -12-56=-68 -14-48=-62 -16-42=-58 -21-32=-53 -24-28=-52$$

The solution is the pair that gives sum -68.

$$p=-56 q=-12$$

Rewrite 32b^{2}-68b+21 as \left(32b^{2}-56b\right)+\left(-12b+21\right).

$$\left(32b^{2}-56b\right)+\left(-12b+21\right)$$

Factor out 8b in the first and -3 in the second group.

$$8b\left(4b-7\right)-3\left(4b-7\right)$$

Factor out common term 4b-7 by using distributive property.

$$\left(4b-7\right)\left(8b-3\right)$$

Evaluate

$$\left(4b-7\right)\left(8b-3\right)$$

Solution Steps

Combine -56b and -12b to get -68b.

$$32b^{2}-68b+21$$