Polynomial Remainder
Consider the function f ( x ) = x 2 0 1 5 + 2 0 1 5 . If f ( x ) is divided by x 8 − x 6 + x 4 − x 2 + 1 , it leaves a remainder g ( x ) . If f ( x ) is divided by ( x + 1 ) 3 , it has a remainder h ( x ) .
Find the value of 1 − g ( − 1 ) h ( 1 ) + 1 .
The answer is 4025.
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3 solutions
Could you please explain the second and third step.
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We have A 5 + 1 = ( A + 1 ) ( A 4 − A 3 + A 2 − A + 1 ) so substitute A = x 2 and you get the first equation. So x 8 − x 6 + x 4 − x 2 + 1 divides x 1 0 + 1 .
B + 1 divides B 2 0 1 + 1 so substitute B = x 1 0 and you get the second equation. So x 1 0 + 1 divides x 2 0 1 0 + 1 .
The third equation basically means x 2 0 1 0 + 1 divides x 2 0 1 5 + x 5 .
So x 8 − x 6 + x 4 − x 2 + 1 divides x 2 0 1 5 + x 5 . From this we get the remainder g ( x ) = − x 5 + 2 0 1 5 .
Also how did you factorise x^10 + 1
Very Nice Problem!
I'm confused by g(x) = -x^5 + 2015. When x=0, we get g(0) = -0^5+2015 = 2015. But g(0) is supposed to be the remainder when we perform f(0)/(x^8-x^6+x^4-x^2+1) = 2015 / 1. Anything divided by 1 has remainder equal to 0. So shouldn't g(0) = 0?
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When we divide polynomials a ( x ) by b ( x ) we have a ( x ) = b ( x ) q ( x ) + r ( x ) where r ( x ) is the remainder polynomial. This division is not limited to integer coefficients. Note that the degree of r ( x ) is less than the degree of b ( x ) but this does not imply that r ( x ) < b ( x ) for all x.
For example, when we divide x 3 + 2 3 x + 1 by 2 x 2 + 1 we have x 3 + 2 3 x + 1 = ( 2 x 2 + 1 ) ( 2 1 x ) + x + 1 so the remainder is x + 1 .
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Thanks, I see that I was missing the quotient, q(0), in my calculation involving the remainder. In your solution, it looks like you derived the remainder without having to get the actual quotient, but it is clear that q(0) would have to be equal to 0, since f(0) = 2015 = q(0)(0^8-0^6+0^4-0^2+1) + g(0) = q(0)(1) + 2015 = q(0) + 2015.
thx bro but for h(x) the other mehod u can go with differential
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Yes, I followed the differential procedure. Very nice question ! P.S. I used Wolfram Alpha for differentiation, I am too lazy
Can u set a hyperlink for your more questions? I am really interested in ur problems.
Prophet's way.
Can U tell me how U got x 1 0 + 1 ∣ ( x 1 0 ) 2 0 1 + 1 ?
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Let y = x 1 0 .
Use induction to prove that (y+1) is a factor of (y^n+1) for integer n.
Not necessarily applied, these are good expressions finally found:
h(x) = -2029105 x^2 - 4056195 x -2025076
g(x) = -x^5 + 2015
h(1) = -8110376
g(-1) = 2016
Therefore, [h (1) + 1]/ [1 - g (-1)] = (-8110375)/ (-2015) = 4025
We should actually do the divisions as these are not one or two factors of polynomial x^2015 + 2015 concerned. The thing is no need to divide all but to extract figure of features for analysis.
x^2015 = (x^8 - x^6 + x^4 - x^2 + 1)(x^2007 + x^2005 - x^1997 - x^1995 + x^1987 + x^1985 ... + x^7 + x^5) - x^5 {For g (-1), Q (-1) is from 2007 down to 5.}
x^2015 = (x^3 + 3 x^2 + 3 x + 1)(x^2012 - 3 x^2011 + 6 x^2010 - 10 x^2009 + 15 x^2008 ... - 2021055 x^3 + 2023066 x^2 - 2025078 x +2027091 - 2029105 x^(-1) + 2031120 x^(-2) - 2033136 x^(-3) ...) {For h (1), Q (1) is from 2012 down to 0.}
For g (-1), substitutes x = -1,
(-1)^2015 = (1 - 1 + 1 - 1 + 1)(-2) + g' (-1)
g' (-1) = 1
(-1)^2015 + 2015 = (1 - 1 + 1 - 1 + 1)(-2) + g (-1)
g (-1) = 2016 {Found 1}
For h (1), substitutes x = 1,
(1)^2015 = (1 + 3 + 3 + 1)(1014049) + h' (1)
h' (1) = 1 - 8112392 =-8112391
(1)^2015 + 2015 = (1 + 3 + 3 + 1)(1014049) + h (1)
h (1) = 1 - 8112392 + 2015 =-8110376 {Found 2}
For h(1), I took (x^3 + 3 x^2 + 3 x + 1)(x^4 - 3 x^3 + 6 x^2 - 10 x + 15) - 21 x^2 - 35 x - 15 for comparison. I found that [c (-1)] x^2 + [3 c (-1) + c (-2)] x + [3 c (-1) + 3 c (-2) + c(-3)] is always the quadratic remainder wanted. Therefore, I obtained the whole expression h (x) rather than just h (1) although not really needed.
Similarly, I noticed (x^8 - x^6 + x^4 - x^2 + 1)(x^7 + x^5) - x^5 by knowing +x^15 is the one to be divided at last for a remainder of -x^5.
The main thing of this question is x^2015 is actually does not have intermediate terms that makes a difficulty while making easy situations with the two dividers, where + 2015 is actually not taking part but just to be added back later.
As both dividers are NOT factor of x^2015 + 2015, any short cut I attempted had first failed. I applied Excel for this question. 4025 is the answer!
x 1 0 + 1 = ( x 2 + 1 ) ( x 8 − x 6 + x 4 − x 2 + 1 )
x 1 0 + 1 ∣ ( x 1 0 ) 2 0 1 + 1 = x 2 0 1 0 + 1
x 2 0 1 0 + 1 ∣ x 5 ( x 2 0 1 0 + 1 ) = x 2 0 1 5 + x 5
Thus x 8 − x 6 + x 4 − x 2 + 1 ∣ x 2 0 1 5 + x 5 .
Since x 2 0 1 5 + 2 0 1 5 = x 2 0 1 5 + x 5 + ( − x 5 + 2 0 1 5 ) thus g ( x ) = − x 5 + 2 0 1 5 . Note that 1 − g ( − 1 ) = − 2 0 1 5 .
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now f ( x ) = x 2 0 1 5 + 2 0 1 5 = ( x + 1 − 1 ) 2 0 1 5 + 2 0 1 5 = ( x + 1 ) 2 0 1 5 − 2 0 1 5 ( x + 1 ) 2 0 1 4 + . . . [terms of ( x + 1 ) 3 or higher] + . . . − 2 2 0 1 5 ⋅ 2 0 1 4 ( x + 1 ) 2 + 2 0 1 5 ( x + 1 ) − 1 + 2 0 1 5 .
Thus it is clear that h ( x ) = − 2 2 0 1 5 ⋅ 2 0 1 4 ( x + 1 ) 2 + 2 0 1 5 ( x + 1 ) + 2 0 1 4 . Now h ( 1 ) + 1 = − 2 0 1 5 ⋅ 2 0 1 4 ⋅ 2 + 2 0 1 5 ⋅ 2 + 2 0 1 5 = − 2 0 1 5 ⋅ 4 0 2 5 .
The answer is thus 4025.