Solve -20m^2q-3mq+35q

$$- 20 m ^ { 2 } q - 3 m q + 35 q$$

Factor

$$-q\left(4m-5\right)\left(5m+7\right)$$

Solution Steps

Factor out q.

$$q\left(-20m^{2}-3m+35\right)$$

Consider -20m^{2}-3m+35. Factor the expression by grouping. First, the expression needs to be rewritten as -20m^{2}+am+bm+35. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-20\times 35=-700$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -700.

$$1,-700 2,-350 4,-175 5,-140 7,-100 10,-70 14,-50 20,-35 25,-28$$

Calculate the sum for each pair.

$$1-700=-699 2-350=-348 4-175=-171 5-140=-135 7-100=-93 10-70=-60 14-50=-36 20-35=-15 25-28=-3$$

The solution is the pair that gives sum -3.

$$a=25 b=-28$$

Rewrite -20m^{2}-3m+35 as \left(-20m^{2}+25m\right)+\left(-28m+35\right).

$$\left(-20m^{2}+25m\right)+\left(-28m+35\right)$$

Factor out -5m in the first and -7 in the second group.

$$-5m\left(4m-5\right)-7\left(4m-5\right)$$

Factor out common term 4m-5 by using distributive property.

$$\left(4m-5\right)\left(-5m-7\right)$$

Rewrite the complete factored expression.

$$q\left(4m-5\right)\left(-5m-7\right)$$

Evaluate

$$-q\left(4m-5\right)\left(5m+7\right)$$