Find $p (0), p (1), p (2)$ for $p (t) = 2 + t + 2 t^{2} + t^{3}$

2, 6, 20

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4, 5, 15

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4, 8, 28

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None of the above

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Solution

The correct option is C $2, 6, 20$
Here, $p (t) = 2 + t + 2 t^{2} + t^{3}$
Now, $p (0) = 2 + 0 + 2 \times 0^{2} + 0^{3} = 2$
$p (1) = 2 + 1 + 2 \times 1^{2} + 1^{3} = 2 + 1 + 2 + 1 = 6$
$p (2) = 2 + 2 + 2 \times 2^{2} + 2^{3} = 2 + 2 + 8 + 8 = 20$

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