# factoring completely

• Apr 11th 2007, 02:48 PM
factoring completely
how do you completely factor 2m (to the 3rd power) minus 5m(squared) plus 3m
• Apr 11th 2007, 03:36 PM
topsquark
Quote:

how do you completely factor 2m (to the 3rd power) minus 5m(squared) plus 3m

2m^3 - 5m^2 + 3m

= m(2m^2 - 5m + 3)

= m(2m - 3)(m - 1)

-Dan
• Apr 11th 2007, 08:00 PM
horsejumper
Quote:

how do you completely factor 2m (to the 3rd power) minus 5m(squared) plus 3m

First take out 1 m so you have a degree of two for you equation
like this
(^2 means to the second power or squared)

m(2m^2-5m+3)

then factor
I used guess and check
I tried several combiations until I figured it out
I tried
m(2m-1)(m-3)
and i came out with
m(2m^2-4m+3)
almost but the middle number was wrong so
I tried
m(2m-3)(m-1)
I checked and got
m(2m^2-5m+3)
bingo!
also
another method of solving is useing the quatdratic formulat

(-b+or- the sqare root (b^2-4AC))/(2A)
so you would fill in
(-(-5) + or - the square root ((-5)^2-(4(2)(3))/((2)(2))

you will get 3/2 and 1
so you can fill them in two ways and both will work
(m-3/2)(m-1)
or
(2m-3)(m-1)

DOnt forget multiplying by the first m you factored out in the begining