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

Last edited:

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! 