- #1
anemone
Gold Member
MHB
POTW Director
- 3,883
- 115
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 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
anemone said:Thanks for participating but I'm sorry, kaliprasad. Your answer is incorrect.
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
A polynomial is an algebraic expression that consists of variables and coefficients, combined using addition, subtraction, and multiplication, but not division or raising to a power. The highest power of the variable in a polynomial is called the degree.
The "Polynomial Challenge" is a mathematical problem that involves finding the sum of the values of a polynomial function at four different inputs, p, q, r, and s. The challenge is to determine the value of f(p)+f(q)+f(r)+f(s) using the given polynomial function, f(x).
To solve the "Polynomial Challenge", you will need to substitute the given values of p, q, r, and s into the polynomial function, f(x), and then add the resulting values together. This will give you the sum of f(p), f(q), f(r), and f(s), which is the solution to the challenge.
Yes, a polynomial can have more than one variable. In fact, polynomials can have any number of variables, as long as the variables are combined using addition, subtraction, and multiplication. For example, the polynomial 2x^2 + 3xy + 5y^2 has two variables, x and y.
Solving polynomial challenges is important as it helps to develop problem-solving skills and strengthens understanding of polynomial functions. It also has real-life applications in fields such as engineering, physics, and economics, where polynomial functions are used to model and solve various problems.