# Polynomial Challenge

#### anemone

##### MHB POTW Director
Staff member
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 $f(x)=x^6-x^5-x^3-x^2-x$.

##### Well-known member
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

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#### anemone

##### MHB POTW Director
Staff member
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

##### Well-known member
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

#### anemone

##### MHB POTW Director
Staff member
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!