The degree of p(x) = x^2(1 + x + x^2) + 5 is. | vedclass

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

Question

(C) To find the degree of the polynomial $p(x) = x^2(1 + x + x^2) + 5$,we first expand the expression.By distributing $x^2$ into the parentheses,we ge

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) To find the degree of the polynomial $p(x) = x^2(1 + x + x^2) + 5$,we first expand the expression.By distributing $x^2$ into the parentheses,we get:$p(x) = x^2(1) + x^2(x) + x^2(x^2) + 5$$p(x) = x^2 + x^3 + x^4 + 5$Rearranging the terms in descending order of their powers,we have:$p(x) = x^4 + x^3 + x^2 + 5$The degree of a polynomial is defined as the highest power of the variable $x$ present in the expression.In this polynomial,the highest power of $x$ is $4$.Therefore,the degree of $p(x)$ is $4$.

The degree of $p(x) = x^2(1 + x + x^2) + 5$ is.............

Similar Questions

Find $k$ such that $x^{2}+2x+k$ is a factor of $2x^{4}+x^{3}-14x^{2}+5x+6$. Also,find all the zeroes of the two polynomials.

There are $x^{2}-2$ students in the class. $x^{3}-3x^{2}+5x-3$ chocolates are distributed between them. Each student should get the maximum possible number of chocolates. Find the number of chocolates received by each student and the number of chocolates left undistributed $(x \in N)$.

$P(x) = 4x^{3} + 3x^{2} + 2x + 1$ is a $\ldots$ polynomial.

Find the division of the following by synthetic division method: $p(x) = x^{4} - a^{4}$ by $x - a$.

Answer the following and justify: Can $x^{2}-1$ be the quotient on division of $x^{6}+2 x^{3}+x-1$ by a polynomial in $x$ of degree $5$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module