#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

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

#2

damn, I really need to review my imaginary numbers

#3

shutup and play your guitar

edit:Try the math thread, someone must be there to help you out.

edit:Try the math thread, someone must be there to help you out.

#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

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.

And if I'm wrong, then that just proves I suck at math.

#6

Shallots!

#7

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.....

*Last edited by zero27 at Oct 23, 2009,*

#8

Look at my edit...

#9

I'm dumb.

*Last edited by wamguy89 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

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

#11

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?

#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.

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

*Last edited by wamguy89 at Oct 23, 2009,*

#13

Did you get it?

#14

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.....

#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

and this is why we have the math thread.There is no way to get 21 in that problem because none of the numbers are multiples of 7... You know?