State the degree of the following polynomial: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To find the degree of the polynomial $p(x) = 5x^8 - 6(x^2)^5 - 4x^3 - 14$,we first simplify the expression.Using the power rule $(x^a)^b = x^{a

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) To find the degree of the polynomial $p(x) = 5x^8 - 6(x^2)^5 - 4x^3 - 14$,we first simplify the expression.Using the power rule $(x^a)^b = x^{a 	imes b}$,we simplify the term $6(x^2)^5$ as $6x^{10}$.Now,the polynomial becomes $p(x) = 5x^8 - 6x^{10} - 4x^3 - 14$.The degree of a polynomial is defined as the highest power of the variable $x$ present in the expression.Comparing the powers of $x$ $(8, 10, 3)$,the highest power is $10$.Therefore,the degree of the polynomial is $10$.

State the degree of the following polynomial: $p(x) = 5x^8 - 6(x^2)^5 - 4x^3 - 14$.

Similar Questions

Examine the validity of the following statement: $(x-2)$ is a factor of $x^{3}+5x^{2}-2x-25$.

The zeros of the quadratic polynomial are $-3$ and $4$. Which of the following represents this polynomial?

Prove that $-6, -\frac{1}{2}$ and $1$ are the zeros of the cubic polynomial $p(x) = 2x^3 + 11x^2 - 7x - 6$. Also,verify the relationship between the zeros and the coefficients.

Find the number of real zeros of the following polynomial: $p(x) = x^{3} - 9x$.

Find the zeroes of the polynomial $x^{2}+\frac{1}{6} x-2,$ and verify the relation between the coefficients and the zeroes of the polynomial.

Vedclass Test Series

Exam Paper Generator

Online Exam Module