Solve 12a^{3}b^2-28a^2b^2-24ab^2

$$12 a ^ { 3 } b ^ { 2 } - 28 a ^ { 2 } b ^ { 2 } - 24 a b ^ { 2 }$$

Factor

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

Solution Steps

Factor out 4.

$$4\left(3a^{3}b^{2}-7a^{2}b^{2}-6ab^{2}\right)$$

Consider 3a^{3}b^{2}-7a^{2}b^{2}-6ab^{2}. Factor out ab^{2}.

$$ab^{2}\left(3a^{2}-7a-6\right)$$

Consider 3a^{2}-7a-6. Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa-6. To find p and q, set up a system to be solved.

$$p+q=-7 pq=3\left(-6\right)=-18$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -7.

$$p=-9 q=2$$

Rewrite 3a^{2}-7a-6 as \left(3a^{2}-9a\right)+\left(2a-6\right).

$$\left(3a^{2}-9a\right)+\left(2a-6\right)$$

Factor out 3a in the first and 2 in the second group.

$$3a\left(a-3\right)+2\left(a-3\right)$$

Factor out common term a-3 by using distributive property.

$$\left(a-3\right)\left(3a+2\right)$$

Rewrite the complete factored expression.

$$4ab^{2}\left(a-3\right)\left(3a+2\right)$$

Evaluate

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