What is the value of $\dfrac{f(-5)+f(9)}{4}$ in the Polynomial Challenge III?

[anemone]

Let $f(x)=x^4+px^3+qx^2+rx+s$, where $p,\,q,\,r,\,s$ are real constants. Suppose $f(3)=2481$, $f(2)=1654$, $f(1)=827$.

Determine the value of $\dfrac{f(-5)+f(9)}{4}$.

 

[mente oscura]

Let $f(x)=x^4+px^3+qx^2+rx+s$, where $p,\,q,\,r,\,s$ are real constants. Suppose $f(3)=2481$, $f(2)=1654$, $f(1)=827$.

Determine the value of $\dfrac{f(-5)+f(9)}{4}$.

Hello.

Spoiler

$$f(1)=827=827*1$$

$$f(2)=1654=827*2$$

$$f(3)=2481=827*3$$

Therefore:

$$\dfrac{f(-5)+f(9)}{4}=\dfrac{827*(-5)+827*(9)}{4}=\dfrac{827(-5+9)}{4}=827$$

Regards

 

[anemone]

Hello.

Spoiler

$$f(1)=827=827*1$$

$$f(2)=1654=827*2$$

$$f(3)=2481=827*3$$

Therefore:

$$\dfrac{f(-5)+f(9)}{4}=\dfrac{827*(-5)+827*(9)}{4}=\dfrac{827(-5+9)}{4}=827$$

Regards

Good try! But that isn't correct...sorry!:)

 

[kaliprasad]

Spoiler

We have f(1) = 827 *1, f(2) = 827*2 , f(3) = 827*3

Hence f(x) – 827 x = 0 for x = 1, 2, or 3

So f(x) = a(x-b)(x-1)(x-2)(x-3) + 827 x where b is the 4th zero of f(x)- 827 x

Comparing coefficient of $x^4$ we have a = 1

so f(x) = (x-b)(x-1)(x-2)(x-3) + 827 x

rest is as below

f(9) = (9-b) * 8 * 7 * 6 + 827 * 9 = 336(9-b) + 827 * 9

f(-5) = (-5-b) * (-6) *(-7) * (-8) + 827 * (-5) = 336(5+b) + 827 * (-5)

So (f(9) + f(-5))/4 = (336 * 14 + 827 * (4))/4 = 2003

 

[anemone]

Spoiler

We have f(1) = 827 *1, f(2) = 827*2 , f(3) = 827*3

Hence f(x) – 827 x = 0 for x = 1, 2, or 3

So f(x) = a(x-b)(x-1)(x-2)(x-3) + 827 x where b is the 4th zero of f(x)- 827 x

Comparing coefficient of $x^4$ we have a = 1

so f(x) = (x-b)(x-1)(x-2)(x-3) + 827 x

rest is as below

f(9) = (9-b) * 8 * 7 * 6 + 827 * 9 = 336(9-b) + 827 * 9

f(-5) = (-5-b) * (-6) *(-7) * (-8) + 827 * (-5) = 336(5+b) + 827 * (-5)

So (f(9) + f(-5))/4 = (336 * 14 + 827 * (4))/4 = 2003

Well done, kaliprasad and thanks for participating!

 

[I like Serena]

Let $f(x)=x^4+px^3+qx^2+rx+s$, where $p,\,q,\,r,\,s$ are real constants. Suppose $f(3)=2481$, $f(2)=1654$, $f(1)=827$.

Determine the value of $\dfrac{f(-5)+f(9)}{4}$.

Spoiler

Define $g(x)=f(x+2)$. Then:
$$g(x)=x^4+Px^3+Qx^2+Rx+S$$
for some real constants $P, Q, R, S$.It follows that:
$$g(0) = S = 1654$$
$$g(1)+g(-1)=(1 + P + Q + R + S) + (1-P+Q-R+S) = 2+2Q+2S = 2481 +827 = 3308$$
$$Q = \frac{3308 - 2 - 2S}{2} = 1653 - S = 1653 - 1654 = -1$$

$$\dfrac{f(-5)+f(9)}{4} = \dfrac{g(-7)+g(7)}{4}
=\frac 1 4 (2\cdot 7^4 + 2 Q\cdot 7^2 + 2S)
=\frac 1 4 (2\cdot 7^4 - 2 \cdot 7^2 + 2\cdot 1654)
=2003$$

 

[anemone]

Spoiler

Define $g(x)=f(x+2)$. Then:
$$g(x)=x^4+Px^3+Qx^2+Rx+S$$
for some real constants $P, Q, R, S$.It follows that:
$$g(0) = S = 1654$$
$$g(1)+g(-1)=(1 + P + Q + R + S) + (1-P+Q-R+S) = 2+2Q+2S = 2481 +827 = 3308$$
$$Q = \frac{3308 - 2 - 2S}{2} = 1653 - S = 1653 - 1654 = -1$$

$$\dfrac{f(-5)+f(9)}{4} = \dfrac{g(-7)+g(7)}{4}
=\frac 1 4 (2\cdot 7^4 + 2 Q\cdot 7^2 + 2S)
=\frac 1 4 (2\cdot 7^4 - 2 \cdot 7^2 + 2\cdot 1654)
=2003$$

Bravo, I like Serena! Hey, your method gives me a very good insight on how to define for another function to relate to the given function and use it to our advantage! Thanks for that and thanks for participating.:)