# 2x^2+9x-5I have been struggling over this factoring problem literally all day. It makes absolutely no sense to me. I break it down into (x )(x ) so now I have to factor -5 right? however -5...

2x^2+9x-5

I have been struggling over this factoring problem literally all day. It makes absolutely no sense to me. I break it down into (x )(x ) so now I have to factor -5 right? however -5 is a prime number and 5 and 1, regardless of where the negative is placed, doesn't seem to add up to 9. Please don't answer this with rows of variables. explain in detail.

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`2x^2+9x-5 = 0`

`2x^2+10x-x-5 = 0`

`2x(x+5)-1(x+5) = 0`

`(x+5)(2x-1) = 0`

**x = -5 or x = 1/2**

Note:

`x^2+bx+c = 0`

When you factoring this kind of function you need to adjust 'b' such a way that b = p+q where pq = c

In this case your question is not like this. it is `2x^2+9x-5 = 0`

To do this you need 1 in front of `x^2` instead of 2. So you can divide the equation by 2 as the first step. This is one method of solving that. There is the point you have failed.

`ax^2+bx+c = 0`

**When you factoring this kind of function you need to adjust 'b' such a way that b = p+q where pq = ac**

In this case your question is like this. it is 2x^2+9x-5 = 0

**So when you are factoring you must find the factors that add up -5*2 = -10 NOT 5.**

This numbers are 10 and-1 which gives addition as 9 and product as -10.

Hope you have understand the issue.

**Sources:**