Find the number of quadratic polynomials ax² + bx + c such that:
a) a, b, c are distinct.
b) a, b, c ε {1, 2, 3, …2008}
c) x + 1 divides ax² + bx + c
Reasoning->b=a+c b ranges from 3 to 2008 x1+x2=n has n-1C1 solns. Using th elogic we get 2+3+..2007 solutions. A total of 2015027 solutions.here,condition 1 comes into picture telling us that a,b,c are distinct. Therefore the even values of b ie.4,6,8,..2008 will have one solution less ie.from the given solution subtract 1003 thus giving us 20104024
Find the number of quadratic polynomials ax² + bx + c
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number-theory
combinatorics
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0if you are copying something from a source, you should link to the source. If you are not satisfied with the proof, you should describe why you aren't satisfied and what would satisfy you. – 2011-05-16
3 Answers
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Ross's answer was my first thought, but as a slightly-different alternative, use the remainder theorem:
The remainder when a polynomial $P(x)$ is divided by $x-k$ is $P(k)$.
So if $x+1=x-(-1)$ divides $P(x)=ax^2+bx+c$ (leaving remainder $0$), $P(-1)=0$, which should give the same condition on $a$, $b$, and $c$ that would result from long division.
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0I have uploaded my answer ......just below the question. – 2011-05-16
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Hint: try polynomial long division to find a condition on $a,b,c$. Then it is a combinatoric question how many sets of that kind there are.
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0@ Ross,please have a look on my logic just below the question – 2011-05-16
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Your question and answers to your question can be found here. For the answer see: abhitsian's Comment in this link:
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0Yes i have taken this question from the site,the problem is i am not satisfied wiith the logic,thats why i wanted a different approach to tackle this question,i also pasted the solution given there but still logic is vauge.the point is i am in search of simple and clear solution.i dont think i am wrong. – 2011-05-16