Solve 14x^2-8+3x-5x^2-12

$$14 x ^ { 2 } - 8 + 3 x - 5 x ^ { 2 } - 12$$

Evaluate

$$\left(3x-4\right)\left(3x+5\right)$$

Solution Steps

Combine 14x^{2} and -5x^{2} to get 9x^{2}.

$$9x^{2}-8+3x-12$$

Subtract 12 from -8 to get -20.

$$9x^{2}-20+3x$$

Factor

$$\left(3x-4\right)\left(3x+5\right)$$

Solution Steps

Multiply and combine like terms.

$$9x^{2}+3x-20$$

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=3 ab=9\left(-20\right)=-180$$

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 -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 3.

$$a=-12 b=15$$

Rewrite 9x^{2}+3x-20 as \left(9x^{2}-12x\right)+\left(15x-20\right).

$$\left(9x^{2}-12x\right)+\left(15x-20\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(3x-4\right)+5\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(3x+5\right)$$