Witryna16 cze 2024 · Click here 👆 to get an answer to your question ️ k is a zero of the polynomial p (x) =x2×2 -11x+24. If k is a prime number then find the value of kwhat … WitrynaTranscribed Image Text: Let ƒ(x) be a polynomial of degree n > 0 in a polynomial ring K[x] over a field K. Prove that any element of the quotient ring K[x]/ (ƒ(x)) is of the form g(x)+(ƒ(x)), where g(x) is a polynomial of degree at most n - 1. Expert Solution. Want to see the full answer?
Answered: Let ƒ(x) be a polynomial of degree n >… bartleby
WitrynaSuppose that $P (x)$ is a polynomial of degree $n$ such that $P (k)=\dfrac {k} {k+1}$ for $k=0,1,\ldots,n$. Find the value of $P (n+1)$ - Mathematics Stack Exchange Suppose that is a polynomial of degree such that for . Find the value of Ask Question Asked 9 years, 2 months ago Modified 9 years, 2 months ago Viewed 8k times 8 WitrynaAnswer (1 of 8): Among the five non-collapsed answers as of writing this: * One correctly answers a totally different question. Namely, “What are examples of a zero degree polynomial?” This one word makes all the difference. * Another incorrectly defines a zero polynomial using the definition ... free tv licence form
Polynomials - What are Polynomials? Definition and Examples
WitrynaSolution Verified by Toppr Correct option is A) Consider the polynomial, p(x)=ax 2+bx+c. To find the zero of a polynomial, we write p(x)=0. Hence, a real number c is said to be a zero of the polynomial p(x), if p(c)=0. Therefore, option A is correct. Solve any question of Polynomials with:- Patterns of problems > Was this answer helpful? 0 0 Witryna9 cze 2015 · Yes: first, note that the derivative of a polynomial $p$ of degree $n$ has degree $n-1$.Let the roots of $p$ be $\alpha_1, \dotsc, \alpha_n$ in increasing order, not necessarily distinct. Further, the factor theorem shows that $$ p (x) = a \prod_k (x-\alpha_k), $$ so $p (x)/a$ is a real-valued polynomial. WitrynaSolution Verified by Toppr Correct option is B) A zero of a polynomial p(x) is a number c such that p(c)=0 Let p(x)=x−3 ∴p(−3)=−3−3=−6 =0 Hence, −3 is not a zero of x−3. So, given statement is false. Solve any question of Polynomials with:- Patterns of problems > Was this answer helpful? 0 0 Similar questions free tv key channel list