Question 173398: Factor the quadratic equation.
3x^2+13x-10
From what I know, one has to find the factors of -10, which are 2 and 5, and 10 and 1, including the negatives. After that, I think one has to find the sums that equal 13x, but I don't see how that would be possible.
Found 2 solutions by Mathtut, solver91311:
Answer by Mathtut(3670)   (Show Source):
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website! You have only considered half of the problem. What you said is true IF the coefficient on the high order term is 1, but here it is 3.
The part about needing the factors of 10 is almost right. In fact you need the factors of -10. So, -2 and 5, 2 and -5, -1 and 10, or 1 and -10.
But the sum required to get the 13 coefficient must consider the 3 coefficient on the high order term.
 , but 
So
Yeah, but how do I know that a quadratic trinomial is factorable in the first place so I don't go rooting (so to speak) around for factors that might not exist? you ask.
Use the discriminant part of the quadratic formula,  where a, b, and c are the coefficients of the 2nd degree, 1st degree, and constant terms respectively. If the number is a perfect square, then the quadratic is factorable, otherwise not. For the given problem,  , so your example is factorable. (If you haven't done the quadratic formula yet, don't worry about it. Just use the rule above on faith for the time being)
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