The degree of p(x) = dots is 5. Which of the | vedclass

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

Question

(C) The degree of a polynomial is the highest power of the variable present in the expression.For option $A$: $x^{3} + x^{2}$,the degree is $3$.For op

Vedclass · Accepted Answer

$x^{3}(x^{2} + 1)$

Vedclass · Answer

(C) The degree of a polynomial is the highest power of the variable present in the expression.For option $A$: $x^{3} + x^{2}$,the degree is $3$.For option $B$: $5x + x^{2}$,the degree is $2$.For option $C$: $x^{3}(x^{2} + 1) = x^{5} + x^{3}$,the degree is $5$.For option $D$: $x(x^{5} - 2) = x^{6} - 2x$,the degree is $6$.Therefore,the polynomial with degree $5$ is $x^{3}(x^{2} + 1)$.

The degree of $p(x) = \dots$ is $5$. Which of the following polynomials satisfies this condition?

Similar Questions

The zeros of $p(x) = 2 + x - x^{2}$ are.......

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 $5$?

If the product of the zeros of $p(x) = ax^{2} - 6x - 6$ is $4$,then $a = $ ............

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

Divide $x^{4}+2x^{3}+3x^{2}+2x+20$ by $x^{2}+2x+2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module