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January 28, 2015

January 28, 2015

Posted by **Aoi** on Monday, June 15, 2009 at 10:09am.

Question:

The cubic polynomial f(x) is such that the coefficient of x^3 is 1 and the roots of f(x)=0 are 1, k, and k^2. IT is given that f(x) has a remainder of 7 when divided by x-2.

Show that k^3 - 2k^2 - 2k - 3 = 0 __________________________________________________________________________

this is what i did, it might be a stupid or completely wrong way of doing..... please pardon my stupidity

x^3 + ax^2 + bx^2 + c = (x-1)(x-k)(x-k^2)

f(2)=7

8+4a+2b+c=7

4a+2b+c=-1

(x-1)(x-k)(x-k^2)= x^3 + (-k-1-k^2)x^2 + (-k^3+k^2+k)x -k^3

so... (-k-1-k^2)= a

(-k^3 + k^2 + k)=b

-k^3 =c

4(-k^2-k-1)+2(-k^3+k^2+k)-k^3= -1

-3k^3-2k^2-2k-3= 0

!!??? which is different from what i;m supposed to show!?!?!?

I honestly dont know what im doing.........

- POLYNOMIALS -
**Aoi**, Monday, June 15, 2009 at 10:24amok... i know how to do this question already! :D

- POLYNOMIALS -
**Aoi**, Monday, June 15, 2009 at 10:28amFor the previous part, this was how i got the answer:

f(x) = (x-1)(x-k)(x-k^2)

= Q(x)(2-x)+7

Let x=2, so (2-k)(2-k^2)=7, and i get the answer,

But there is another part to this question which i dont understand...

please help:D

it says: Hence, find a value of k and show that there are no other real values of k which satisfy this equation.

How do i do this??

- POLYNOMIALS -
**Reiny**, Monday, June 15, 2009 at 12:02pmk^3 - 2k^2 - 2k - 3 = 0

after a few quick trial & error guesses

I found k=3 to work

so by division I got

k^3 - 2k^2 - 2k - 3 = 0

(k-3)(k^2 + k + 1) = 0

so any other solutions for k must come from

k^2 + k + 1=0

Using the formula it can be quickly shown that there is no "real" solution to this

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