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fp1 complex numbers

https://b4be22b3d902bfcdbf6f1c96474d2c6cea0afe43.googledrive.com/host/0B1ZiqBksUHNYalpOMnd5c1kwRnc/Edexcel-Set-2/Ch.1%20Complex%20Numbers.pdf

I get 2x^2 -2x+13 with a remainder of 36 when i divide f(x) by x-3

what can i do to go further?

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Reply 1
Avatar for JLegion
JLegion
2
Original post by timebent
https://b4be22b3d902bfcdbf6f1c96474d2c6cea0afe43.googledrive.com/host/0B1ZiqBksUHNYalpOMnd5c1kwRnc/Edexcel-Set-2/Ch.1%20Complex%20Numbers.pdf

I get 2x^2 -2x+13 with a remainder of 36 when i divide f(x) by x-3

what can i do to go further?


You shouldn't be getting a remainder.

Did you do so by long division? You have likely made an error. If you can't spot it feel free to post a picture of your workings for us to look at :smile:
Reply 2
Avatar for timebent
timebent
OP
2
Original post by JLegion
You shouldn't be getting a remainder.

Did you do so by long division? You have likely made an error. If you can't spot it feel free to post a picture of your workings for us to look at :smile:


-.-' eugh lemme try again maybe my long division's rusty
funned up the last bit haha
(edited 7 years ago)
Reply 3
Avatar for B_9710
B_9710
18
Remainder is 25 not 36.
Reply 4
Avatar for timebent
timebent
OP
2
Original post by B_9710
Remainder is 25 not 36.


no remainders....
Reply 5
Avatar for JLegion
JLegion
2
Original post by B_9710
Remainder is 25 not 36.


There is no remainder.
Reply 6
Avatar for timebent
timebent
OP
2
Original post by JLegion
You shouldn't be getting a remainder.

Did you do so by long division? You have likely made an error. If you can't spot it feel free to post a picture of your workings for us to look at :smile:


I got as answers x=3 x=1±32x= \dfrac{1\pm \sqrt 3}{2}
Reply 7
Avatar for B_9710
B_9710
18
Original post by JLegion
There is no remainder.
Original post by timebent
no remainders....
I thought the question was divide 2x2−2x+13 2x^2-2x+13 by x−3 x-3 , I can't see the question.
(edited 7 years ago)
Reply 8
Avatar for timebent
timebent
OP
2
Original post by B_9710
I thought the question was divide 2x2−2x+13 2x^2-2x+13 by x−3 x-3 , I can't see the question.


2x3−8x2+7x−3 2x^3 -8x^2 +7x-3 was original question ^^ was what i got the first time i did long division wrong
Reply 9
Avatar for B_9710
B_9710
18
Original post by timebent
2x3−8x2+7x−3 2x^3 -8x^2 +7x-3 was original question ^^ was what i got the first time i did long division wrong


Yes I see now.
Reply 10
Avatar for JLegion
JLegion
2
Original post by timebent
I got as answers x=3 x=1±32x= \dfrac{1\pm \sqrt 3}{2}


I think you should check your workings for mistakes; I got x = 3, 1/2 +/- i/2
Reply 11
Avatar for B_9710
B_9710
18
Original post by timebent
I got as answers x=3 x=1±32x= \dfrac{1\pm \sqrt 3}{2}


Equation has non real roots.
Reply 12
Avatar for timebent
timebent
OP
2
Original post by JLegion
I think you should check your workings for mistakes; I got x = 3, 1/2 +/- i/2


ok since i can't see where i went wrong

here's my workings

(x−3)(2x2−2x−1)(x-3)(2x^2 -2x-1)

−b±b2−4ac2a \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

2±22−4x2x−14\dfrac{2 \pm \sqrt{2^2-4x2x-1}}{4}

2±124 \dfrac{2 \pm \sqrt{12}}{4}

2±234 \dfrac{2 \pm 2\sqrt{3}}{4}

1±32 \dfrac{1 \pm \sqrt{3}}{2}
Reply 13
Avatar for B_9710
B_9710
18
Original post by timebent
ok since i can't see where i went wrong

here's my workings

(x−3)(2x2−2x−1)(x-3)(2x^2 -2x-1)

−b±b2−4ac2a \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

2±22−4x2x−14\dfrac{2 \pm \sqrt{2^2-4x2x-1}}{4}

2±124 \dfrac{2 \pm \sqrt{12}}{4}

2±234 \dfrac{2 \pm 2\sqrt{3}}{4}

1±32 \dfrac{1 \pm \sqrt{3}}{2}


Should get 2x2−2x+1 2x^2-2x+1 as quotient.
Original post by timebent
I got as answers x=3 x=1±32x= \dfrac{1\pm \sqrt 3}{2}


That's close, sorry I was late to see this.

ImageUploadedByStudent Room1469128321.532462.jpg


Posted from TSR Mobile
Anyone, is there a faster way?

It's meant to be part (c)
image.jpeg356.3KB
image.jpeg419.5KB
Reply 16
Avatar for B_9710
B_9710
18
Original post by Chittesh14
Anyone, is there a faster way?

It's meant to be part (c)


It's very painful for me to read when you don't use exact values and use decimal approximations.
Original post by B_9710
It's very painful for me to read when you don't use exact values and use decimal approximations.


Hmm.. Don't worry then, I don't think any other way will be much faster.


Posted from TSR Mobile
Reply 18
Avatar for B_9710
B_9710
18
Original post by Chittesh14
Hmm.. Don't worry then, I don't think any other way will be much faster.


Posted from TSR Mobile


Probably not.
Original post by B_9710
Probably not.


:smile:. Just need to go over chapter 1 before I continue because I don't think I remember everything.


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