Find the value of $p(x) = 2x^3 + x^2 - x - 1$ at $x = 0$ and $x = -2$.
(N/A) Given polynomial: $p(x) = 2x^3 + x^2 - x - 1$
Step $1$: Find the value at $x = 0$.
$p(0) = 2(0)^3 + (0)^2 - (0) - 1$
$p(0) = 0 + 0 - 0 - 1$
$p(0) = -1$
Step $2$: Find the value at $x = -2$.
$p(-2) = 2(-2)^3 + (-2)^2 - (-2) - 1$
$p(-2) = 2(-8) + 4 + 2 - 1$
$p(-2) = -16 + 4 + 2 - 1$
$p(-2) = -11$
Thus,the values are $p(0) = -1$ and $p(-2) = -11$.
Step $1$: Find the value at $x = 0$.
$p(0) = 2(0)^3 + (0)^2 - (0) - 1$
$p(0) = 0 + 0 - 0 - 1$
$p(0) = -1$
Step $2$: Find the value at $x = -2$.
$p(-2) = 2(-2)^3 + (-2)^2 - (-2) - 1$
$p(-2) = 2(-8) + 4 + 2 - 1$
$p(-2) = -16 + 4 + 2 - 1$
$p(-2) = -11$
Thus,the values are $p(0) = -1$ and $p(-2) = -11$.
Explore More
Similar Questions
Divide $x^{3}-6x^{2}+11x-6$ by $x^{2}-8x+27$.
Difficult
View SolutionFind the zeroes of the following polynomial by the factorisation method and verify the relationship between the zeroes and the coefficients of the polynomial: $4x^2 - 3x - 1$
Medium
View SolutionThe number of real zeros of $y=p(x)$ is ............... in the given figure.
Easy
View SolutionIdentify the type of the given polynomial based on its degree: $p(x) = (x - 1)(2 - x)$
Easy
View SolutionObtain a quadratic polynomial with the following conditions:
The sum of the zeros $= -\frac{1}{4}$;
The product of the zeros $= \frac{1}{4}$.
The sum of the zeros $= -\frac{1}{4}$;
The product of the zeros $= \frac{1}{4}$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo