From the following figure,find the number of zeros of $y=p(x)$ :

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(0) The number of zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $X-$axis.
In the given figure,the graph of $y=p(x)$ lies entirely above the $X-$axis and does not intersect or touch the $X-$axis at any point.
Therefore,the number of real zeros of the polynomial $p(x)$ is $0$.

Explore More

Similar Questions

The zeros of the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$; where $a \neq 0$ and $a, b, c, d \in R$,are $\alpha, \beta$,and $\gamma$. Then $\alpha + \beta + \gamma = \ldots$

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

The number of real zeros of $y = p(x)$ is $\ldots \ldots \ldots$ in the given figure.

Divide $14x^3 - 5x^2 + 9x - 1$ by $2x - 1$.

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

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo