Solve 3a^2+4a-15a-20

$$3 a ^ { 2 } + 4 a - 15 a - 20$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$3a^{2}-11a-20$$

Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa-20. To find p and q, set up a system to be solved.

$$p+q=-11 pq=3\left(-20\right)=-60$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -11.

$$p=-15 q=4$$

Rewrite 3a^{2}-11a-20 as \left(3a^{2}-15a\right)+\left(4a-20\right).

$$\left(3a^{2}-15a\right)+\left(4a-20\right)$$

Factor out 3a in the first and 4 in the second group.

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

Factor out common term a-5 by using distributive property.

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

Evaluate

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

Solution Steps

Combine 4a and -15a to get -11a.

$$3a^{2}-11a-20$$