Solve 12r^2s^2+7rs^2-10s^2

$$12 r ^ { 2 } s ^ { 2 } + 7 r s ^ { 2 } - 10 s ^ { 2 }$$

Factor

$$\left(3r-2\right)\left(4r+5\right)s^{2}$$

Solution Steps

Factor out s^{2}.

$$s^{2}\left(12r^{2}+7r-10\right)$$

Consider 12r^{2}+7r-10. Factor the expression by grouping. First, the expression needs to be rewritten as 12r^{2}+ar+br-10. To find a and b, set up a system to be solved.

$$a+b=7 ab=12\left(-10\right)=-120$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 7.

$$a=-8 b=15$$

Rewrite 12r^{2}+7r-10 as \left(12r^{2}-8r\right)+\left(15r-10\right).

$$\left(12r^{2}-8r\right)+\left(15r-10\right)$$

Factor out 4r in the first and 5 in the second group.

$$4r\left(3r-2\right)+5\left(3r-2\right)$$

Factor out common term 3r-2 by using distributive property.

$$\left(3r-2\right)\left(4r+5\right)$$

Rewrite the complete factored expression.

$$s^{2}\left(3r-2\right)\left(4r+5\right)$$

Evaluate

$$\left(3r-2\right)\left(4r+5\right)s^{2}$$