# confused while trying to factor this

• Oct 30th 2010, 02:55 PM
NecroWinter
confused while trying to factor this
y^3-y^2+3y-3 is what im factoring

the book says the answer is (y-1)(y^2+3)

i dont know how it gets there

heres what i try to do:
(y^3-y^2)+(3y-3)
-y^2(y-1) + 3(y-1)

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

what am i doing wrong?
as usual, thanks for any help. this site really helps me out
• Oct 30th 2010, 03:09 PM
harish21
Quote:

Originally Posted by NecroWinter
y^3-y^2+3y-3 is what im factoring

the book says the answer is (y-1)(y^2+3)

i dont know how it gets there

heres what i try to do:
(y^3-y^2)+(3y-3)
-y^2(y-1) + 3(y-1) : the common factor for the first term should be y^2.

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

what am i doing wrong?
as usual, thanks for any help. this site really helps me out

Look at the line in red. You error is there.

\$\displaystyle (y^3-y^2)+(3y-3) \$

\$\displaystyle =y^2(y-1)+3(y-1) = (y-1)(y^2+3)\$
• Oct 30th 2010, 03:33 PM
NecroWinter
ok now that leads me to a new question

my book has me factoring out negative numbers as well sometimes, how do you know what youre supposed to do? is it just when a negative number is all the way to the left?
• Oct 30th 2010, 03:51 PM
harish21
You can take a negative factor here as well.

\$\displaystyle (y^3-y^2)+(3y-3)\$

\$\displaystyle =-y^2(-y+1)-3(-y+1) = (-y+1)(-y^2-3)\$

\$\displaystyle =(-y+1)(-y^2-3)\$

\$\displaystyle =-(y-1)[-(y^2+3)] = (y-1)(y^2+3)\$

You can follow whichever way you think is convenient.
• Oct 30th 2010, 03:51 PM
harish21
double posted by mistake..