It is as hard as it doesn't seems!!

Algebra Level 4

Let $f(x)$ be a 5th degree polynomial with $f(1)=1$ , $f(2)=2$ , $f(3)=3$ , $f (4)=4$ , $f(5)=5$ , and $f(6)=36$ . Find the value of $f(10)-10$ .

The answer is 3780.

2 solutions

Chew-Seong Cheong
Mar 3, 2018

From the fact that $f(n) = n$ for $n = 1,2,3,4,5$ , the 5th degree polynomial can be of the form

$\begin{aligned} f(x) & = {\color{#3D99F6}a}(x-1)(x-2)(x-3)(x-4)(x-5)+x & \small \color{#3D99F6} \text{where }a \text{ is a constant.} \\ \implies f(6) & = a(5)(4)(3)(2)(1) + 6 = 120a+6 = 36 & \small \color{#3D99F6} \implies a = \frac 14 \\ \implies f(x) & = \frac 14 (x-1)(x-2)(x-3)(x-4)(x-5) + x \\ f(10) - 10 & = \frac 14 (9)(8)(7)(6)(5) + 10 - 10 \\ & = \boxed{3780} \end{aligned}$

Thanks for making my solution more clear

Aryan Sanghi - 3 years, 3 months ago

You are welcome. You can such enclose your formula with "\ (" and "\ )" (no space between "\" and "(" and "\" and ")" to put the formula in LaTex You can place your mouse cursor over the formulas to see the LaTex codes or click the pull-down menu " $\cdots$ More" and select "Toggle LaTex" to see the LaTex codes.

Chew-Seong Cheong - 3 years, 3 months ago

Thanks for explaining me LaTex

Aryan Sanghi - 3 years, 3 months ago

Sir, can you explain how the general form of a fifth degree polynomial is like that? How would it be of a sixth degree polynomial and so on ?

Aman thegreat - 3 years, 3 months ago

From the fact that $f(n)=n$ for $n=1,2,3,4,5$ . We note that $f(x) = a(x-1)(x-2)(x-3)(x-4)(x-5) + x$ provides solution for $x=n$ , because for $x=1,2,3,4,5$ , $a(x-1)(x-2)(x-3)(x-4)(x-5)=0$ , then $f(x) = x$ . If you have 6th degree you will need another $f(x_1) = b$ . So that we can solve for the unknown $a$ . In general to solution an $n$ th degree polynomial of that form we need $n+1$ conditions or equations.

Chew-Seong Cheong - 3 years, 3 months ago

Aryan Sanghi
Mar 2, 2018

The above 5th degree equation can be written as f(x)=k(x-1)(x-2)(x-3)(x-4)(x-5)+x

Where k is a constant.

Putting x=6, we get k= $\frac{1}{4}$

Now, putting x=10, we get f(10)-10=3780

It is as hard as it doesn't seems!!

The answer is 3780.

2 solutions

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