Answer: t-9 and t+11
The order doesn't matter, so you could say "t+11 and t-9" to mean the same thing.
=======================================================
Explanation:
Think of two numbers that multiply to -99, but also add to 2. In other words, find two factors of -99 which add to 2. We'll use trial and error.
Here are all the ways to multiply to -99 using integers only:
- 1 and -99
- 3 and -33
- 9 and -11
- 11 and -9
- 33 and -3
- 99 and -1
Add up each factor pair to see which sum is equal to 2.
- 1 + (-99) = -98
- 3 + (-33) = -30
- 9 + (-11) = -2
- 11 + (-9) = 2 is what we want
- 33 + (-3) = 30
- 99 + (-1) = 98
We can see that 11 and -9 are the two numbers we're after, so therefore, t^2+2t-99 factors to (t-9)(t+11). The order doesn't matter with the factor multiplication.
As a more efficient alternative method, you can use the quadratic formula to solve t^2+2t-99 = 0 to end up with t = 9 and t = -11. That would lead to the two equations t-9 = 0 and t+11 = 0. Applying the zero product property then gets us (t-9)(t+11) = 0.
To help show that (t-9)(t+11) becomes t^2+2t-99, you can use the FOIL rule to expand that factorization out. This is one way to verify the answer.
