Posted by **poof** on Friday, March 7, 2008 at 12:07pm.

i am having a hard time with factoring i have read what is here but am still having trouble can some one help me with the problem s^2-s-6, and y^2-4y-5. these are expressions which either can be factored or are prime. can someone explain how to do these in regular terms.

- math -
**drwls**, Friday, March 7, 2008 at 12:34pm
They factor into (s-3)(s+2) and

(y-5)(y+1)

Look at the constant term at the end and see what prime number it can be factored into. In the case of 6 (in the first problem), it can be either 3 and 2, or 6 and 1. Because of the minus sign in front of the 6, you have to have opposite signs on the two numbers that you pick. Also, the sum of the two must be -1 (the coefficient of the middle "s" term). The numbers that meet these requirements are -3 and +2. Hence the factors are (s-3) and (s+2)

If the coefficent of the "square" term were not 1, the problem gets a bit more complex, but it still comes down to looking for prime number factors of the squared term's coefficient and the constant term, computing the middle term, and a bit of trial and error.

- math -
**Joshua**, Friday, March 7, 2008 at 12:37pm
s^2 - s - 6

What you're looking for is a number that when you multiply it, it equals

-6 and when you add them it equals -1. these values -6 and -1 are derived from the equation.

so number could be --> -/+ 3, -/+ 2

but since there's only one way of getting -1 when you add them. the answer is -3 and 2.

when you multiply -3 and 2==> -6

when you add them it equal ==> -1

so ==> ans= (s-3)(s+2)

Try to do the same for the other problem.

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