#1
ill post a quick question here since no one is in the math help thread at all

how would you get (x-(2+i)) (x-(2-i)) (x-4)

to x^3 - 8x^2 + 21x - 20

when i multiply it out i get x^3 -9x^2 + 20x + 20
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#4
I got almost the answer: x^3-8x^2+17x-4

EDIT: Here's my work...

(x-2-i)(x-2+i)=

x^2 - 2x + ix - 2x + 4 - 2i - ix + 2i - i^2 =

x^2 - 4x + 5

Then multiply that by the (x-4)

x^3 - 8x^2 + 21x - 20
Last edited by wamguy89 at Oct 23, 2009,
#5
I can't believe I'm helping with math, but I think the PEMDAS trick might help.


And if I'm wrong, then that just proves I suck at math.
#7
Quote by rnrsoldier1461
ill post a quick question here since no one is in the math help thread at all

how would you get (x-(2+i)) (x-(2-i)) (x-4)

to x^3 - 8x^2 + 21x - 20

when i multiply it out i get x^3 -9x^2 + 20x + 20

should you have wrote that problem as....

[x-(2+i)] [x-(2-i)] (x-4)

?

I haven't done this type of math in a long time but i got (-2x* + 8x) i used the * as a symbol for squared since I don't know how to type the little 2 .... I'm not 100% on that answer though I'm curious now though.

edit: to show my work i did....

[x-(2+i)] [x-(2-i)] (x-4)
(-2x-xi) (-2+xi) (x-4)
-2x(x-4)
-2x*+8x

I don't know man.....
If you want to shine like the sun first you must burn like it.
Last edited by zero27 at Oct 23, 2009,
#10
watch your negatives!

you can rewrite as
(x-2-i)(x-2+i)(x-4)

(x^2 - 2x + ix - 2x + 4 - 2i - ix + 2i -(i^2))(x-4)
remember that i^2 = -1 so it becomes

(x^2 - 2x + ix - 2x + 4 - 2i - ix + 2i +1)(x-4)

simplifying,
(x^2 - 4x +5)(x-4)

x^2 - 4x^2 + 5x - 4x^2 + 16x - 20

simplifying again,
x^3 - 8x^2 +21x - 20

simple...i'm sure you didn't watch your negatives correctly. Good luck
Quote by crazydiamond73
i killed a hooker while she was servicing me and we were both high on crack, all on the teachers desk. i mean c'mon.
#11
Quote by wamguy89
There is no way to get 21 in that problem because none of the numbers are multiples of 7... You know?


you're kidding me right?
Quote by crazydiamond73
i killed a hooker while she was servicing me and we were both high on crack, all on the teachers desk. i mean c'mon.
#12
Oh wow... I just forgot to add my 4 and 1 to get 5, but I would have had it right, otherwise... I edited it, though...

EDIT: zero27 is very wrong... Don't look at his work.
Last edited by wamguy89 at Oct 23, 2009,
#14
Quote by wamguy89
Oh wow... I just forgot to add my 4 and 1 to get 5, but I would have had it right, otherwise... I edited it, though...

EDIT: zero27 is very wrong... Don't look at his work.

thanks alot jerk.....
If you want to shine like the sun first you must burn like it.
#15
Haha sorry man! I just didn't want him to do that! It's really bothering me that you wrote "alot" instead of "a lot," too... Sorry!!!! I feel bad now... I'm a terrible person. My humblest apologies and a hug!
Last edited by wamguy89 at Oct 23, 2009,
#16
Quote by wamguy89
There is no way to get 21 in that problem because none of the numbers are multiples of 7... You know?
and this is why we have the math thread.
Sent from my iPad.