Find $p(0)$,$p(1)$,and $p(2)$ for the following polynomial: $p(t) = 2 + t + 2t^2 - t^3$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Given the polynomial: $p(t) = 2 + t + 2t^2 - t^3$.
To find $p(0)$,substitute $t = 0$ into the polynomial:
$p(0) = 2 + (0) + 2(0)^2 - (0)^3 = 2 + 0 + 0 - 0 = 2$.
To find $p(1)$,substitute $t = 1$ into the polynomial:
$p(1) = 2 + (1) + 2(1)^2 - (1)^3 = 2 + 1 + 2(1) - 1 = 2 + 1 + 2 - 1 = 4$.
To find $p(2)$,substitute $t = 2$ into the polynomial:
$p(2) = 2 + (2) + 2(2)^2 - (2)^3 = 2 + 2 + 2(4) - 8 = 2 + 2 + 8 - 8 = 4$.

Explore More

Similar Questions

Give possible expressions for the length and breadth of each of the following rectangles,in which their areas are given: $\text{Area} = 35y^2 + 13y - 12$.

Factorise $6x^2 + 17x + 5$ by splitting the middle term,and by using the Factor Theorem.

Difficult
View Solution

Verify whether the following is a zero of the polynomial indicated against it:
$p(x) = 5x - \pi, \, x = \frac{4}{5}$

Divide $p(x)$ by $g(x)$,where $p(x) = x + 3x^2 - 1$ and $g(x) = 1 + x$.

Find the degree of the polynomial given: $x^{5}-x^{4}+3$

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