If the value of y =1 then find the value of $2y^3 + 3y^2 + y - 3$.

Given: 

Value of $y$ =1

To find:

Value of $2y^3 + 3y^2 + y - 3$.

Solution:

$2y^3 + 3y^2 + y - 3$

Substitute $y$ = 1,

$2(1)^3 +3(1)^2+ 1 - 3$  

$2 (1) + 3 (1)  - 2  =  2 + 3 - 2  =  5 - 2  =  3$

 The value of $2y^3 + 3y^2 + y - 3$ at y=1 is 3.

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Updated on: 10-Oct-2022

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