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Actually, in such questions where a polynomial $f(x)$ follows a pattern for certain values in its domain, we generally use a trick like defining a new polynomial $g(x)$ whose roots are known, and then calculating $f(x)$. For instance, here we could have done this by defining $g(x) = f(x) - \frac{1}{x+1}$. Then the roots of $g(x)$ are $1,2,3,4$. Then, $g(x) = k(x-1)(x-2)(x-3)(x-4)$, giving us $f(x) = k(x-1)(x-2)(x-3)(x-4) + \frac{1}{x+1}$. But then, the degree of $f(x)$ will be $4$, but we are given its degree to be $3$. And in such cases, I think we have only one option, i.e. to use Lagrange Interpolation.

@Raushan Sharma
–
No,this question can still be done by Remainder Factor Theorem.

Let $g(x)=(x+1)f(x)-1$ You have to express it in this form in order to make $g(x)$ a polynomial.Anyways, $g(x)=0$ for $x=1,2,3,4$.So $(x+1)f(x)-1=g(x)=c(x-1)(x-2)(x-3)(x-4)$ for some constant $c$.Putting $x=-1$ we get:
$-1=c(-2)(-3)(-4)(-5)\implies c=\frac{-1}{120}$
Therefore $g(x)=(x+1)f(x)-1=\frac{-1}{120}(x-1)(x-2)(x-3)(x-4)$ Now simply putting $x=5$ gives:
$6\times f(5)-1=\frac{-1}{120} (4)(3)(2)(1)=\frac{-1}{5}\\
\implies \boxed{f(5)=\frac{2}{15}}$

@Abdur Rehman Zahid
–
Ohh, that's really nice. So, this method also works here.... Thanks for your solution (+1). And definitely it's better than Interpolation in such cases, where there is a pattern given. Thanks once again!! :D

@Abdur Rehman Zahid
–
Yes, but then, $f(x) = g(x) + \frac{1}{x+1}$ is not a polynomial. So, there is some flaw in this method. That's why I didn't do it with this method, as I said earlier. A traditional method of Lagrange Interpolation, as I mentioned earlier is the best here, I think.

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## Comments

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TopNewestThe answer should be $\frac{2}{15}$. Use Lagrange Interpolation

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Can u solve more easily? Or Please explain your solution!

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Actually, in such questions where a polynomial $f(x)$ follows a pattern for certain values in its domain, we generally use a trick like defining a new polynomial $g(x)$ whose roots are known, and then calculating $f(x)$. For instance, here we could have done this by defining $g(x) = f(x) - \frac{1}{x+1}$. Then the roots of $g(x)$ are $1,2,3,4$. Then, $g(x) = k(x-1)(x-2)(x-3)(x-4)$, giving us $f(x) = k(x-1)(x-2)(x-3)(x-4) + \frac{1}{x+1}$. But then, the degree of $f(x)$ will be $4$, but we are given its degree to be $3$. And in such cases, I think we have only one option, i.e. to use Lagrange Interpolation.

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$g(x)=f(x)-\dfrac{1}{x+1}$ is not a polynomial

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$g(x)$ has roots $1,2,3,4$. So, it can be expressed in the form $g(x) = k(x-1)(x-2)(x-3)(x-4)$, which is a polynomial.

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Let $g(x)=(x+1)f(x)-1$ You have to express it in this form in order to make $g(x)$ a polynomial.Anyways, $g(x)=0$ for $x=1,2,3,4$.So $(x+1)f(x)-1=g(x)=c(x-1)(x-2)(x-3)(x-4)$ for some constant $c$.Putting $x=-1$ we get: $-1=c(-2)(-3)(-4)(-5)\implies c=\frac{-1}{120}$ Therefore $g(x)=(x+1)f(x)-1=\frac{-1}{120}(x-1)(x-2)(x-3)(x-4)$ Now simply putting $x=5$ gives: $6\times f(5)-1=\frac{-1}{120} (4)(3)(2)(1)=\frac{-1}{5}\\ \implies \boxed{f(5)=\frac{2}{15}}$

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$f(x) = g(x) + \frac{1}{x+1}$ is not a polynomial. So, there is some flaw in this method. That's why I didn't do it with this method, as I said earlier. A traditional method of Lagrange Interpolation, as I mentioned earlier is the best here, I think.

Yes, but then,Log in to reply

Since you are evaluating for consecutive integers you could have used method of differences.

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