Question

solve by factoring
t^2+3t-18=0

Answer 1

The general form of polynomials like this are (Xn +/- Constant1) * (Yn +/- Constant2)

Since the constant in the equation (-18) is negative, one of the constants in our new equation has to be negative, and other has to positive. And since our t^2 term has no coefficient, neither X nor Y can be anything besides 1. So rewriting:

(t + Constant1) * (t - Constant2)

Now we have to be clever. We need to find out which factors of 18 would ever add or subtract to be the coefficient of the t term. The factors of 18 are 1, 2, 3, 6, 9 and 18. We notice that 6 minus 3 would equal 3 so we put 6 in the positive spot, and 3 minus spot.

(t + 6)(t - 3) = 0

Answer 2

The answer is (t+6)(t-3)=0

To check it you can use distributive property to check your answer