State the degree of the given polynomial: p(x | vedclass

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

Question

(B) The given polynomial is $p(x) = x^{2} - \sqrt{3}x^{3} + 4x^{7} + 9$.To find the degree,we first arrange the terms in descending order of their exp

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The given polynomial is $p(x) = x^{2} - \sqrt{3}x^{3} + 4x^{7} + 9$.To find the degree,we first arrange the terms in descending order of their exponents:$p(x) = 4x^{7} - \sqrt{3}x^{3} + x^{2} + 9$.The degree of a polynomial is defined as the highest power of the variable present in the polynomial.In this expression,the powers of $x$ are $7, 3, 2,$ and $0$ (since $9 = 9x^{0}$).The maximum power is $7$.Therefore,the degree of the polynomial is $7$.

State the degree of the given polynomial: $p(x) = x^{2} - \sqrt{3}x^{3} + 4x^{7} + 9$.

Similar Questions

The zero of $p(x) = -x^2 + 2x - 1$ is..........

The zeros of $p(x) = 3x^2 - x - 4$ are $\alpha$ and $\beta$,then $\alpha^2 \beta + \alpha \beta^2 = \ldots$

Find the division of the following by synthetic division method: $p(t) = 2t^2 + 3t + 1$ by $t + 2$.

The value of $p(x) = 2x^{4} - 3x^{3} + 7x + 5$ at $x = -2$ is $\ldots \ldots \ldots \ldots$

Find a quadratic polynomial,the sum and product of whose zeroes are $\sqrt{2}$ and $-\frac{3}{2},$ respectively. Also,find its zeroes.

Vedclass Test Series

Exam Paper Generator

Online Exam Module