Dividing a Polynomial of Degree 2 0 1 3
If the polynomial x 2 0 1 3 leaves a remainder p x 2 + q x + r when divided by x 3 − x , then find the value of p , q and r .
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3 solutions
I was confused when it said x 2 0 1 3 , thought it meant there was stuff after it. It should say If x 2 0 1 3 without the word polynomial
x 3 − x x 2 0 1 3 − x 2 0 1 1 + x 3 − x x 2 0 1 1
once again
= x 2 0 1 0 + x 3 − x x 2 0 1 1 − x 2 0 0 9
when we will go on doing this we see that always x 2 n remains
thus the smallest power of x which x 3 − x would divide is 4
thus by long division method x 3 − x x 4 we get remainder as x
thus p = r = 0 and q =1
by Long division,
(x^2013 + 0x^2012 + ... + 0)/(x^3 - x) = (x^2010 + 0x^2009 + ... + 0) r. 0x^2 + 1x + 0
At first...
(x^2013 + 0x^2012 + 0x^2011)/(x^3 - x) = x^2010 r. 0x^2012 + 1x^2011 +
(0x^2012 + 1x^2011 + 0x^2010)/(x^3 - x) = 1x^2011 + 0x^2010
So, the pattern continues,
it will end to.
0x^2 + 1x + 0
and that's the remainder. :D
And substitute it and getting p=0 , q=1 , r=0.
x 2 0 1 3 = ( x 3 − x ) f ( x ) + p x 2 + q x + r .
If x = 0 , r = 0
If x = − 1 , p − q = − 1 and if x = 1 , p + q = 1 .
So, p = 0 , q = 1