2n^2+7n+6

[Guest]

okay so im starting college and i have no idea why when i solve 2n^2+7n+6 i get (x+2) (x+3/2). When tested these two numbers make the equation 0 but if you make the products you get (2n^2+7n+6)/2, so in theory the roots would be (n+2) (2n+3), but why, can som1 explain it to me? xD ( i know is a silly question )

[Guest]

[tex]2n^2+7n+6 = (n+2)(2n+3)[/tex]

Roots are -2 and -3/2

-3 canot be a root because 2n is -3 but n=-3/2

[HallsofIvy]

okay so im starting college and i have no idea why when i solve 2n^2+7n+6 i get (x+2) (x+3/2).

To begin with, I don't know what you mean when you say "solve 2n^2+ 7n+ 6"! Do you mean "solve the equation 2n^2+ 7n+ 6= 0"?

And then I don't know why you would say "I get (x+ 2)(x+ 3/2)"! That has the variable "x" while your original expression has "n". Even (n+ 2)(n+ 3/2)= n^2+ 2n+ (3/2)n+ 6/2= n^2+ (7/2)n+ 3= (1/2)(2n^2+ 7n+ 6), not "2n^2+ 7n+ 6"

What you can say is that 2n^2+ 7n+ 6= 2(n+ 2)(n+ 3/2). If the problem is to solve 2n^2+ 7n+ 6= 0 then from 2n^2+ 7n+ 6= 2(n+ 2)(n+ 3/2)= 0 you can divide both sides by 2 to get (n+ 2)(n+ 3/2)= 0. Now 0 has two very nice properties: If either a= 0 or b= 0 (or both) then ab= 0 and if ab= 0 at least one of a or b (or both) is 0. So from (n+ 2)(n+ 3/2)= 0 either n+ 2= 0 or n+ 3/2= 0 (or both). If n+ 2= 0 then n= -2 and if n+ 3/2= 0 then n= -3/2.

When tested these two numbers make the equation 0 but if you make the products you get (2n^2+7n+6)/2, so in theory the roots would be (n+2) (2n+3), but why, can som1 explain it to me? xD ( i know is a silly question )

Yes, If I were doing this I would have factored 2n^2+ 7n+ 6 as (n+ 2)(2n+ 3). Again, if the problem were 2n^2+ 7n+ 6= 0 from (n+ 2)(2n+ 3)= 0 we have either n+ 2= 0 or 2n+ 3= 0 (or both). From n+ 2= 0, n= -2 again and from 2n+ 3= 0 we have 2n= -3 and then n= -3/2 again.