Question

if p(x)= 3y^3 +y^2 +2y+1 . find the value of p(0) p(2).​

Answer 1

Answer:

p(0) = 0+0+0+1 = 1

p(2) = 24+4+4+1 = 33

Step-by-step explanation:

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Answer 2

AnsweR :-

$p(y) = {3y}^{3} + {y}^{2} + 2y + 1$

By putting 0 in place of y then :-

$p(0) = {3(0)}^{2} + {(0)}^{2} + 2(0) + 1$

$p(0) = 0 + 0 + 0 + 1$

$p(0) = 1$

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By putting 2 in place of y then:-

$p(2) = {3(2)}^{3} + {(2)}^{2} + 2(2) + 1$

$p(2) = {24}^{} + {4}^{} + 4+ 1$

$p(2) = 33$

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Procedure:-

$\implies$ In First equation we will put 0 in place of y and then we will equate it. As you can see that first one is 0, then we know that whatever number we will multiply with 0 will become 0 so, the first three will become zero then added by one.

$\implies$ In the second one we will place 2 in place of y. As yiyu can see that y as power 3 and 2. First we will multiply 2 by itself 3 times it will come 8 then it will be multiply by 3 so it will be 24 same with raise to power 2.

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