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Dillinger4
Jul 25th, 2003, 02:36 PM
Im not quite sure if this is how this is commonly solved. Any help would be great.

I have to factor. y(y+6 ) - 3(y + 6) which ends up (y +6)(y-3).

Would this be the correct way to factor?

y2 + 6y - 3y - 18
y2 + 3y - 18

Now since y * y = y2 I put (y (y and since the L in FOIL has to equal -18 i end up with (y + 6)(y-3) bucause 6 * -3 = 18.

DiGiTaIErRoR
Jul 25th, 2003, 02:51 PM
You did it correctly.

Dillinger4
Jul 25th, 2003, 03:01 PM
Thanks for the help. :)

sql_lall
Jul 26th, 2003, 04:54 AM
yeah, thats right.

in general, if you have:

y2 + ay + b

and you want to factorise to:

(y+g)(y+h)

then you just need to find the two numbers such that:
g+h = a
gh = b

i.e. their sum is a, and their product is b.

Guv
Jul 26th, 2003, 08:19 PM
Looks like more work than necessary.

I have to factor. y(y+6 ) - 3(y + 6) which ends up (y +6)(y-3).

Suppose you were asked to factor: y*x - 3*x?

Wouldn't you immediately reply: (y - 3) *x ?

Think of (y + 6) as x.