Solve 3a^2-6a+4a-8=

$$3 a ^ { 2 } - 6 a + 4 a - 8 =$$

Factor

$$\left(a-2\right)\left(3a+4\right)$$

Solution Steps

Multiply and combine like terms.

$$3a^{2}-2a-8$$

Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa-8. To find p and q, set up a system to be solved.

$$p+q=-2 pq=3\left(-8\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -2.

$$p=-6 q=4$$

Rewrite 3a^{2}-2a-8 as \left(3a^{2}-6a\right)+\left(4a-8\right).

$$\left(3a^{2}-6a\right)+\left(4a-8\right)$$

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

$$3a\left(a-2\right)+4\left(a-2\right)$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(3a+4\right)$$

Evaluate

$$\left(a-2\right)\left(3a+4\right)$$

Solution Steps

Combine -6a and 4a to get -2a.

$$3a^{2}-2a-8$$