MHB Polynomial Challenge: Find $f(p)+f(q)+f(r)+f(s)$

AI Thread Summary
The discussion revolves around finding the value of $f(p)+f(q)+f(r)+f(s)$ for the polynomial roots of $x^4-x^3-x^2-1=0$, where $f(x)=x^6-x^5-x^3-x^2-x$. Participants initially provided incorrect answers, but corrections were made, leading to a successful solution. The final consensus confirms that the correct solution was achieved. The interaction highlights the collaborative effort in problem-solving within the forum.
anemone
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The roots of $x^4-x^3-x^2-1=0$ are $p, q, r, s$. Find $f(p)+f(q)+f(r)+f(s)$, where $f(x)=x^6-x^5-x^3-x^2-x$.
 
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X^4 = x^3+x^2 +1
So x^6 = x^5 + x^4 + x^2
So x^6 – x^5 – x^3 – x^2 – x =x^4-x^3 – x = x^2- x + 1

So f(x) = x^2
Sum p = 1 and sum pq = 0

Sum p^2 =( sum p)^2 – 2 sum pq = 1
sum p = 1

so sum p^2 - p + 1 = 1
 
Last edited:
kaliprasad said:
X^4 = x^3+x^2 +1
So x^6 = x^5 + x^4 + x^2
So x^6 – x^5 – x^3 – x^2 – x =x^4-x^3 – x = x^2- x + 1

So f(x) = x^2
Sum p = 1 and sum pq = 0

Sum p^2 =( sum p)^2 – 2 sum pq = 1
sum p = 1

so sum p^2 - p + 1 = 1

Thanks for participating but I'm sorry, kaliprasad. Your answer is incorrect.
 
anemone said:
Thanks for participating but I'm sorry, kaliprasad. Your answer is incorrect.

Thanks anemone. My ans is incorrect. Here is the correct solution

We are given that(say the function is g)

g(x) = x^4-x^3 –x^2 – 1

As p,q,r,s are 4 roots we have
g(x) = 0 has solution p,q,r,s

Now x^4 – x^3 – x^2 – 1 = 0 ... (1) has solutions p,q,r,s
F(x) = x^6 – x^5 – x^3 – x^2 – x ..(2)

We need to reduce it t the lowest oder polynomial as possible

From (1) x^6 – x^5 = x^4 + x^2 ... (3)

From (2) and (3) F(x) = x^4 + x^2 – x^3 – x^2 – x = (x^3 + x^2 + 1) + x^2 – x^3 – x^2 – x
= x^2 – x + 1

So f(p) + f(q) + f(r) + f(s) = p^2 + q^2 + r^2 + s^2) – (p+q+r+s) + 4 ... (4)
Now as p q r s are roots of x^4 –x^3 – x^2 – x = 0
So using vietas formula
P + q + r + s = 1 ...(5) (
Pq + pr +ps + qr + qs + rs = - 1 ..(6)
Now p^2 + q^2 + r^2 + s^2 = (p+q+r+s)^2 – 2(pq+ Pq + pr +ps + qr + qs + rs) = 1 + 2 = 3 ... (7)

Using (5) and (7) in (4) we get

f(p) + f(q) + f(r) + f(s) = p^2 + q^2 + r^2 + s^2) – (p+q+r+s) + 4 = 3 – 1 + 4 = 6

I hope that solution is correct
 
kaliprasad said:
Thanks anemone. My ans is incorrect. Here is the correct solution

We are given that(say the function is g)

g(x) = x^4-x^3 –x^2 – 1

As p,q,r,s are 4 roots we have
g(x) = 0 has solution p,q,r,s

Now x^4 – x^3 – x^2 – 1 = 0 ... (1) has solutions p,q,r,s
F(x) = x^6 – x^5 – x^3 – x^2 – x ..(2)

We need to reduce it t the lowest oder polynomial as possible

From (1) x^6 – x^5 = x^4 + x^2 ... (3)

From (2) and (3) F(x) = x^4 + x^2 – x^3 – x^2 – x = (x^3 + x^2 + 1) + x^2 – x^3 – x^2 – x
= x^2 – x + 1

So f(p) + f(q) + f(r) + f(s) = p^2 + q^2 + r^2 + s^2) – (p+q+r+s) + 4 ... (4)
Now as p q r s are roots of x^4 –x^3 – x^2 – x = 0
So using vietas formula
P + q + r + s = 1 ...(5) (
Pq + pr +ps + qr + qs + rs = - 1 ..(6)
Now p^2 + q^2 + r^2 + s^2 = (p+q+r+s)^2 – 2(pq+ Pq + pr +ps + qr + qs + rs) = 1 + 2 = 3 ... (7)

Using (5) and (7) in (4) we get

f(p) + f(q) + f(r) + f(s) = p^2 + q^2 + r^2 + s^2) – (p+q+r+s) + 4 = 3 – 1 + 4 = 6

I hope that solution is correct

Yeah, it's correct now! Well done, kali!:cool:
 
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