Solve 12a^2+30a-14a-35

$$12 a ^ { 2 } + 30 a - 14 a - 35$$

Factor

$$\left(6a-7\right)\left(2a+5\right)$$

Solution Steps

Multiply and combine like terms.

$$12a^{2}+16a-35$$

Factor the expression by grouping. First, the expression needs to be rewritten as 12a^{2}+pa+qa-35. To find p and q, set up a system to be solved.

$$p+q=16 pq=12\left(-35\right)=-420$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -420.

$$-1,420 -2,210 -3,140 -4,105 -5,84 -6,70 -7,60 -10,42 -12,35 -14,30 -15,28 -20,21$$

Calculate the sum for each pair.

$$-1+420=419 -2+210=208 -3+140=137 -4+105=101 -5+84=79 -6+70=64 -7+60=53 -10+42=32 -12+35=23 -14+30=16 -15+28=13 -20+21=1$$

The solution is the pair that gives sum 16.

$$p=-14 q=30$$

Rewrite 12a^{2}+16a-35 as \left(12a^{2}-14a\right)+\left(30a-35\right).

$$\left(12a^{2}-14a\right)+\left(30a-35\right)$$

Factor out 2a in the first and 5 in the second group.

$$2a\left(6a-7\right)+5\left(6a-7\right)$$

Factor out common term 6a-7 by using distributive property.

$$\left(6a-7\right)\left(2a+5\right)$$

Evaluate

$$\left(6a-7\right)\left(2a+5\right)$$

Solution Steps

Combine 30a and -14a to get 16a.

$$12a^{2}+16a-35$$