Solve a^2+7a-4a-28=

$$a ^ { 2 } + 7 a - 4 a - 28 =$$

Factor

$$\left(a-4\right)\left(a+7\right)$$

Solution Steps

Multiply and combine like terms.

$$a^{2}+3a-28$$

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-28. To find p and q, set up a system to be solved.

$$p+q=3 pq=1\left(-28\right)=-28$$

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 -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$p=-4 q=7$$

Rewrite a^{2}+3a-28 as \left(a^{2}-4a\right)+\left(7a-28\right).

$$\left(a^{2}-4a\right)+\left(7a-28\right)$$

Factor out a in the first and 7 in the second group.

$$a\left(a-4\right)+7\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(a+7\right)$$

Evaluate

$$\left(a-4\right)\left(a+7\right)$$

Solution Steps

Combine 7a and -4a to get 3a.

$$a^{2}+3a-28$$