Question
If $p(y) = 4 - 3y -y^2 + 5y^3$, find:
$i. p(0)$
$ii. p(2)$
$iii. p(-1)$
$i. p(0)$
$ii. p(2)$
$iii. p(-1)$
✓
Answer
$p(y) = 4 - 3y -y^2 + 5y^3$
$\Rightarrow p(0) = (4 + 3 \times 0 - 0^2 + 5 \times 0^3 )$
$= (4 + 0 - 0 + 0)$
$= 4$
$p(y) = 4 - 3y -y^2 + 5y^3$
$\Rightarrow p(2) = (4 + 3 \times 2 - 2^2 + 5 \times 2^3 )$
$= (4 + 6 - 4 + 40)$
$= 46$
$p(y) = 4 - 3y -y^2 + 5y^3$
$\Rightarrow p(-1) = [(4 + 3 \times (- 1)^2 + 5 \times (-1)^3 )]$
$= (4 - 3 - 1 - 5)$
$= -5$
$\Rightarrow p(0) = (4 + 3 \times 0 - 0^2 + 5 \times 0^3 )$
$= (4 + 0 - 0 + 0)$
$= 4$
$p(y) = 4 - 3y -y^2 + 5y^3$
$\Rightarrow p(2) = (4 + 3 \times 2 - 2^2 + 5 \times 2^3 )$
$= (4 + 6 - 4 + 40)$
$= 46$
$p(y) = 4 - 3y -y^2 + 5y^3$
$\Rightarrow p(-1) = [(4 + 3 \times (- 1)^2 + 5 \times (-1)^3 )]$
$= (4 - 3 - 1 - 5)$
$= -5$
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