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Problem : Suppose $f(x)$ is a polynomial of degree $5$, and with leading coefficient $2009$. If further that $f(1) =1; f(2)=3, f(3)=5, f(4)=7, f(5)=9$. What is the value of $f(6)?$

My work :

Let $f(x) = ax^5+bx^4+cx^3+dx^2+ex+f$

Now $$\begin{align} f(1) &= 2009 \cdot1 +b +c+d+e+f = 1 . . . .(i)\\ f(2) &= 2009\cdot(2)^5+b(2)^4+c(2)^3+d(2)^2+2e+f....(ii)\\ f(3)& = 2009\cdot(3)^5+b(3)^4+c(3)^3+d(3)^2+3e+f....(iii)\\ f(4) &= 2009\cdot(4)^5+b(4)^4+c(4)^3+d(4)^2+4e+f....(iv)\\ f(5) &= 2009\cdot(5)^5+b(5)^4+c(5)^3+d(5)^2+5e+f....(v)\end{align}$$

Can we do this way by solving these five equations or is their any other easier way of doing this.. Please suggest thanks a lot..

choco_addicted
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Sachin
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  • I personally can't see any "easier" way of doing this, but this information gives you five linear equations in $b, c, d, e$, and $f$, and you can solve that by applying Gaussian elimination to the resultant matrix you find. – Alex Wertheim May 10 '13 at 03:08

1 Answers1

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Consider the polynomial $h(x) = f(x)-(2x-1)$ which is also of degree $5$. You are given that $1,2,3,4,5$ are all the $5$ roots of $h$, and that it's leading coefficient is $2009$. Therefore,

$$h(x)=2009(x-1)(x-2)(x-3)(x-4)(x-5)$$

Now putting in $x=6$, $h(6)=2009*5*4*3*2*1=241080$.

This gives $f(6)=h(6)+(2\times 6-1)=241080+11=241091$

Milind Hegde
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