SOLUTION: 2y^2+17y-9 when i find the common factors and write out the equation which i wrote down, (1y+9)(2y1-1) i know it isnt correct. i just have a feeling i am missing a vital and ve

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: 2y^2+17y-9 when i find the common factors and write out the equation which i wrote down, (1y+9)(2y1-1) i know it isnt correct. i just have a feeling i am missing a vital and ve      Log On


   

Question 879337: 2y^2+17y-9
when i find the common factors and write out the equation which i wrote down, (1y+9)(2y1-1) i know it isnt correct.
i just have a feeling i am missing a vital and very simple part of finding the answer. please help
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
What you suggest you want is to factorize 2y%5E2%2B17y-9. This is an expression, but NOT an equation. There are a few combinations of numbers to try, and maybe one of them will work; but only one of them at most, or maybe none of them.

(a+h)(b-k) is supposed to produce the constant term of -9. These would be possible binomial combinations to check:

(2y-1)(y+9)
(2y+1)(y-9)
(2y-9)(y+1)
(2y+1)(y-9)
(2y+3)(y-3)
(2y-3)(y+3)

They all will produce the -9, but which one if any will reproduce the +17y?
.
.
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In fact, the first one listed gives you this:
highlight_green%28%282y-1%29%28y%2B9%29%29---This factorization works.
%282y-1%29y%2B%282y-1%29%2A9
2y%5E2-y%2B18y-9
2y%5E2%2B18y-y-9
2y%5E2%2B%2818-1%29y-9
highlight_green%282y%5E2%2B17y-9%29