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In summary, a polynomial is an algebraic expression involving variables and coefficients, with the highest power of the variable being the degree. The "Polynomial Challenge" is a problem that requires finding the sum of a polynomial function at four different inputs. To solve the challenge, the given values are substituted into the function and added together. Polynomials can have more than one variable, and solving the challenges helps to develop problem-solving skills and has real-life applications in various fields.

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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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kaliprasad

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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:

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anemone

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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 + 1So f(x) = x^2

Sum p = 1 and sum pq = 0Sum p^2 =( sum p)^2 – 2 sum pq = 1

sum p = 1so sum p^2 - p + 1 = 1

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

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kaliprasad

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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

- #5

anemone

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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,sNow 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 + 1So 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**!

## Related to Polynomial Challenge: Find $f(p)+f(q)+f(r)+f(s)$

## 1. What is a polynomial?

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.

## 2. What is the "Polynomial Challenge"?

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).

## 3. How can I solve the "Polynomial Challenge"?

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.

## 4. Can a polynomial have more than one variable?

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.

## 5. What is the importance of solving polynomial challenges?

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.

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