Solve 7p^2q^2-140p^2q+588p^2

$$7 p ^ { 2 } q ^ { 2 } - 140 p ^ { 2 } q + 588 p ^ { 2 }$$

Factor

$$7\left(q-14\right)\left(q-6\right)p^{2}$$

Solution Steps

Factor out 7.

$$7\left(p^{2}q^{2}-20p^{2}q+84p^{2}\right)$$

Consider p^{2}q^{2}-20p^{2}q+84p^{2}. Factor out p^{2}.

$$p^{2}\left(q^{2}-20q+84\right)$$

Consider q^{2}-20q+84. Factor the expression by grouping. First, the expression needs to be rewritten as q^{2}+aq+bq+84. To find a and b, set up a system to be solved.

$$a+b=-20 ab=1\times 84=84$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 84.

$$-1,-84 -2,-42 -3,-28 -4,-21 -6,-14 -7,-12$$

Calculate the sum for each pair.

$$-1-84=-85 -2-42=-44 -3-28=-31 -4-21=-25 -6-14=-20 -7-12=-19$$

The solution is the pair that gives sum -20.

$$a=-14 b=-6$$

Rewrite q^{2}-20q+84 as \left(q^{2}-14q\right)+\left(-6q+84\right).

$$\left(q^{2}-14q\right)+\left(-6q+84\right)$$

Factor out q in the first and -6 in the second group.

$$q\left(q-14\right)-6\left(q-14\right)$$

Factor out common term q-14 by using distributive property.

$$\left(q-14\right)\left(q-6\right)$$

Rewrite the complete factored expression.

$$7p^{2}\left(q-14\right)\left(q-6\right)$$

Evaluate

$$7\left(q-14\right)\left(q-6\right)p^{2}$$