Solve 40a^3b-44a^2b+12ab

$$40 a ^ { 3 } b - 44 a ^ { 2 } b + 12 a b$$

Factor

$$4ab\left(2a-1\right)\left(5a-3\right)$$

Solution Steps

Factor out 4.

$$4\left(10a^{3}b-11a^{2}b+3ab\right)$$

Consider 10a^{3}b-11a^{2}b+3ab. Factor out ab.

$$ab\left(10a^{2}-11a+3\right)$$

Consider 10a^{2}-11a+3. Factor the expression by grouping. First, the expression needs to be rewritten as 10a^{2}+pa+qa+3. To find p and q, set up a system to be solved.

$$p+q=-11 pq=10\times 3=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -11.

$$p=-6 q=-5$$

Rewrite 10a^{2}-11a+3 as \left(10a^{2}-6a\right)+\left(-5a+3\right).

$$\left(10a^{2}-6a\right)+\left(-5a+3\right)$$

Factor out 2a in the first and -1 in the second group.

$$2a\left(5a-3\right)-\left(5a-3\right)$$

Factor out common term 5a-3 by using distributive property.

$$\left(5a-3\right)\left(2a-1\right)$$

Rewrite the complete factored expression.

$$4ab\left(5a-3\right)\left(2a-1\right)$$

Evaluate

$$4ab\left(2a-1\right)\left(5a-3\right)$$