Solve 15a^{2}b-21a^2b^3+6a^2b^2

$$15 a ^ { 2 } b - 21 a ^ { 2 } b ^ { 3 } + 6 a ^ { 2 } b ^ { 2 }$$

Factor

$$3b\left(1-b\right)\left(7b+5\right)a^{2}$$

Solution Steps

Factor out 3.

$$3\left(5a^{2}b-7a^{2}b^{3}+2a^{2}b^{2}\right)$$

Consider 5a^{2}b-7a^{2}b^{3}+2a^{2}b^{2}. Factor out a^{2}b.

$$a^{2}b\left(5-7b^{2}+2b\right)$$

Consider 5-7b^{2}+2b. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-7b^{2}+2b+5$$

Factor the expression by grouping. First, the expression needs to be rewritten as -7b^{2}+pb+qb+5. To find p and q, set up a system to be solved.

$$p+q=2 pq=-7\times 5=-35$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$p=7 q=-5$$

Rewrite -7b^{2}+2b+5 as \left(-7b^{2}+7b\right)+\left(-5b+5\right).

$$\left(-7b^{2}+7b\right)+\left(-5b+5\right)$$

Factor out 7b in the first and 5 in the second group.

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

Factor out common term -b+1 by using distributive property.

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

Rewrite the complete factored expression.

$$3a^{2}b\left(-b+1\right)\left(7b+5\right)$$

Evaluate

$$3b\left(1-b\right)\left(7b+5\right)a^{2}$$