Question

find p(0),p(1) and p(2) for p(t) =2+t+2t²-8t³
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Answer 1

★ A N S W E R :

➡️ Given -

• p (t) = 2 + t + 2t² - 8t³

➡️ To Find -

• p (0) = ?
• p (1) = ?

➡️ Solution -

$\mapsto \rm \: p(0) = 2 + (0) + 2( {0)}^{2} - 8( {0}^{3} )$

$\mapsto \rm \: p(0) = 2 + 0 + 0 - 0$

$\mapsto \boxed{\rm{p(0) = 2}}$

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$\rm \mapsto \: p(2) = 2 + (2) + 2( {2)}^{2} - 8( {2)}^{3}$

$\mapsto \rm \: p(2) = 2 + 2 + 2 \times 4 - 8 \times 8$

$\mapsto \rm \: p(2) = 2 + 2 + 8 - 64$

$\mapsto \boxed{ \rm{p(2) = - 52}}$

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Answer 2

${\huge{\fcolorbox{lime}{black}{\color{yellow}{Answer}}}}$

★ p(t) = 2 + t + 2t² - 8t³

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# p(0) = 2 + 0 + 2×0² - 8×0³

→ p(0) = 2 + 0 + 0 - 0

→ p(0) = 2

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# p(2) = 2 + 2 + 2×2² - 8×2³

→ p(2) = 4 + 8 - 64

→ p(2) = -52

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