Solve 18x^2-15x+48x-40

$$18 x ^ { 2 } - 15 x + 48 x - 40$$

Factor

$$\left(6x-5\right)\left(3x+8\right)$$

Solution Steps

Multiply and combine like terms.

$$18x^{2}+33x-40$$

Factor the expression by grouping. First, the expression needs to be rewritten as 18x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=33 ab=18\left(-40\right)=-720$$

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 -720.

$$-1,720 -2,360 -3,240 -4,180 -5,144 -6,120 -8,90 -9,80 -10,72 -12,60 -15,48 -16,45 -18,40 -20,36 -24,30$$

Calculate the sum for each pair.

$$-1+720=719 -2+360=358 -3+240=237 -4+180=176 -5+144=139 -6+120=114 -8+90=82 -9+80=71 -10+72=62 -12+60=48 -15+48=33 -16+45=29 -18+40=22 -20+36=16 -24+30=6$$

The solution is the pair that gives sum 33.

$$a=-15 b=48$$

Rewrite 18x^{2}+33x-40 as \left(18x^{2}-15x\right)+\left(48x-40\right).

$$\left(18x^{2}-15x\right)+\left(48x-40\right)$$

Factor out 3x in the first and 8 in the second group.

$$3x\left(6x-5\right)+8\left(6x-5\right)$$

Factor out common term 6x-5 by using distributive property.

$$\left(6x-5\right)\left(3x+8\right)$$

Evaluate

$$\left(6x-5\right)\left(3x+8\right)$$

Solution Steps

Combine -15x and 48x to get 33x.

$$18x^{2}+33x-40$$