Question

Factorize \(2a^3-5a^2-10a+2\)

Answer 1

Eh...Looks wrong..:D is the question right? @ujjwal

Answer 2

Its 2a^3.. Just appears to be 2a^2 .. lol

Answer 3

ohh..

Answer 4

\(\huge 2a^3-5a^2-10a+2\)

Answer 5

Thanks jdoe0001

Answer 6

guess you will have to find roots using cardon's method or something

Answer 7

can't squeeze much out of that, unless you use the "rational root test" to get all roots
but dunno if it's required for this exercise

Answer 8

Any idea? Its like,
if f(a)=0,
(x-a) is a factor
but how do i know the value of that a?
can't simply go on testing all integers and fractions out there.. There should be certain values which can be tested or something.. but i don't know how to find that..

Answer 9

http://www.youtube.com/watch?v=GTx-rXTQg1o

Answer 10

here none of the methods would work, this is not simple

Answer 11

both \(a_n\) and \(a_0\) are 2 in this case.. :\

Answer 12

yea so what, that just gives the possible rational roots as : \(\pm 1, \pm 1/2, \pm2\)
none of them would work, so rational root theorem is not useful here

Answer 13

yeah, none of them works..

Answer 14

You could use fundamental theorem of algebra also, and use long division.

Answer 15

what's that?

Answer 16

The beginning, you start by finding a number that is a factor of the polynomial. So, for instance, you pick a few numbers for "x". Say for instance, 0, 1, 2, 3, -1, -3, -4, etc....

Once you have that value that makes the statement true, meaning 0 = 0
you know that the number you chose is a factor of the polynomial. So you have
x-n divided by \(ax^3+bx^2+cx+e\)

Answer 17

and the problem comes because we have a lot of possibilities for x including fractions..
So, that doesn't seem to be a good idea..

Answer 18

Well, have you tried quadratic formula to get a value which u can use to substutue

Answer 19

What you want to do is eliminate to get a simpler polynomial.

Answer 20

It has fractional factors.

Answer 21

\(\frac{5}{6}\)