Solve 5a^2b-10ab-40b

$$5 a ^ { 2 } b - 10 a b - 40 b$$

Factor

$$5b\left(a-4\right)\left(a+2\right)$$

Solution Steps

Factor out 5.

$$5\left(a^{2}b-2ab-8b\right)$$

Consider a^{2}b-2ab-8b. Factor out b.

$$b\left(a^{2}-2a-8\right)$$

Consider a^{2}-2a-8. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-8. To find p and q, set up a system to be solved.

$$p+q=-2 pq=1\left(-8\right)=-8$$

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 -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -2.

$$p=-4 q=2$$

Rewrite a^{2}-2a-8 as \left(a^{2}-4a\right)+\left(2a-8\right).

$$\left(a^{2}-4a\right)+\left(2a-8\right)$$

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

$$a\left(a-4\right)+2\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(a+2\right)$$

Rewrite the complete factored expression.

$$5b\left(a-4\right)\left(a+2\right)$$

Evaluate

$$5b\left(a-4\right)\left(a+2\right)$$