Question

Anyone know how to simplify this?
(x^3-3x+1-3)/(x-2)

Answer 1

I need to group it correct?

Answer 2

see if you can factor the numerator..

Answer 3

Im just blanking on factoring trinomials for some reason, I remember there is a way to group them.

Answer 4

I am able to factor one with X^2 but the cube is throwing me off

Answer 5

it's
\[x ^{3} -3x -2\]
so test the factors of -2 to see which one makes the term equal zero

Answer 6

ok can i do this ?
(x^2-1) (x-2)

Answer 7

not quite

Answer 8

how'd you get the (x^2-1) factor...?

Answer 9

well, if it was only x^2-3x-2 it would be (x-2)(x-1), right?

Answer 10

no it wouldnt my bad

Answer 11

no here's the way to do it.
1. find the factors of -2 they are 1, -1, 2 and -2

Answer 12

test them in x^3-3x-2
to see which is a root

Answer 13

1 isn't 1^3 -3*1 -2 is not equal to zero

Answer 14

-1 is

Answer 15

and 2 is

Answer 16

so (x-2) is a factor... and that means the other factor is going to be (x^2+1)

Answer 17

2 is?

Answer 18

2^3 -3*2 -2 =0

Answer 19

ahhh

Answer 20

(-1)^3 -3*(-1) -2 = 0 also

Answer 21

why does the x^2 go with the -1 and not the 2?

Answer 22

so either (x+1) or (x-2) is a possible factor...

Answer 23

can you just choose for convience of cancelling later?

Answer 24

no.

Answer 25

if you try to use (x+1) you'll see it isn't going to work because the other factor would have to be (x-2)^2 ... and you won't be able to get x^3-3x-2 then

Answer 26

so (x-2)(x+1)^2

Answer 27

ah ok, thanks again !!!! @Algebraic!