Polynomial Challenge

[anemone]

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 $f(x)=x^6-x^5-x^3-x^2-x$.

 

[kaliprasad]

Spoiler

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

 

[anemone]

Spoiler

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.

 

[kaliprasad]

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

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

Spoiler

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

 

[anemone]

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

Spoiler

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!