What do you get when you factor 2q^2+22q=-60?

Is there more than one answer possible?
I got 2 answers:

1. (2q+10)(q+6)
2. (2q+12)(q+5)

Your two answers are both:

(2)(q+6)(q+5)

on a test, if i put one of the answers i put above, would i still get it right?

You should factor out the GCF.

2q^2 + 22q = -60
2q^2 + 22q + 60 = 0
2(q^2 + 11q + 30) = 0

now divide both sides by 2

q^2 + 11q + 30 = 0
(q + 6)(q + 5) = 0

if 2 numbers multiplied together = 0 then one or both factors must be 0.

q + 6 = 0 and/or q + 5 = 0
q = -6 and/or q = -5

Plug in the 2 possible answers and see if one or both of them work.

would i still get it right if i didn't factor out the gcf?

To factor the quadratic equation 2q^2 + 22q = -60, we first set the equation equal to zero by adding 60 to both sides:

2q^2 + 22q + 60 = 0.

Since the leading coefficient is not 1, we cannot use the method of factoring by grouping. Instead, we can use the method of factoring by decomposition.

Here's how you can solve it step by step:

1. Multiply the coefficient of the squared term by the constant term: 2 * 60 = 120.

2. Find two numbers whose product is 120 and sum is equal to the coefficient of the linear term (22). In this case, the numbers are 10 and 12.

3. Rewrite the linear term 22q as the sum of these two numbers:
2q^2 + 10q + 12q + 60 = 0.

4. Group the quadratic terms together and the linear terms together:
(2q^2 + 10q) + (12q + 60) = 0.

5. Factor out the greatest common factor from each grouping:
2q(q + 5) + 12(q + 5) = 0.

6. Notice that we have a common binomial factor (q + 5):
(q + 5)(2q + 12) = 0.

7. Now, simplify the second binomial:
(q + 5)(2(q + 6)) = 0.

So, the factored form of the quadratic equation 2q^2 + 22q = -60 is:
(q + 5)(2q + 12) = 0.

Therefore, you have correctly found the two possible answers:
1. (2q + 10)(q + 6)
2. (2q + 12)(q + 5).