1. ## More Factoring

2y^2 -7y +3

Up until now I've been pretty adept at coming up with the necessary numbers for the middle monomial but for some reason this one and similiar ones like 3y^2 +5y -2 are killing me. I'm pretty sure this is an easy fix but what method should I use?

2. Originally Posted by Ingersoll
2y^2 -7y +3

Up until now I've been pretty adept at coming up with the necessary numbers for the middle monomial but for some reason this one and similiar ones like 3y^2 +5y -2 are killing me. I'm pretty sure this is an easy fix but what method should I use?

Assuming we can factor just using integers, there are four possibilities to begin with, just based on the coefficient 2 in 2y^2.

(2y + u) (y + v)
(2y + u) (y - v)
(2y - u) (y + v)
(2y - u) (y - v)

Seeing the + in front of the three, we can narrow down to

(2y + u) (y + v)
(2y - u) (y - v)

Seeing the - in front of the 7y, we can narrow down to

(2y - u) (y - v)

possible (u,v) are (1,3), (3,1), giving

(2y - 1) (y - 3)
(2y - 3) (y - 1)

A little examination reveals

(2y - 1) (y - 3)

In practice you can do most of the steps in your head, I would just write

Code:
(2y -    )(y -    )
and then fill in the blanks.

3. ## A simple method !

[PHP][/PHP]There is a quite simple method for this equation ! $\displaystyle \ \ ax^2+bx+c \ \$

You have to find two numbers so that there multiplication become $\displaystyle \ \ a\cdot c \ \$ and there sum is like $\displaystyle \ \ b \ \$ for example:

$\displaystyle \ 2y^2-7y+3 \ \ \ we \ can \ see \ that (-1,-6) is \ the \ suitable \ numbers \ so \ that \ \ 2y^2-6y-y+3 \ \ become \ \ 2y(y-3)+(y-3) = (y-3)(2y+1) \ \$

and $\displaystyle 3y^2 +5y -2 \ (-1,6) \ become \ 3y^2+6y-y-2 \ and \ become \ 3y(y+2)-(y+2) = (y+2)(3y-1) \ \ \$