If $p (x) = x^{3} + 3 x^{2} - 2 x + 4$ then find $p (1)$ and $p (- 2)$

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Solution

Find the value of $P$
Given that $P (x) = x^{3} + 3 x^{2} - 2 x + 4$
$\begin{array}{rcl} P (- 2) & = & {(- 2)}^{3} + 3 {(- 2)}^{2} - 2 (- 2) + 4 \\ = & - 8 + 12 + 4 + 4 \\ = & 12 \end{array}$
$\begin{array}{rcl} P (1) & = & {(1)}^{3} + 3 {(1)}^{2} - 2 (1) + 4 \\ = & 1 + 3 - 2 + 4 \\ = & 6 \end{array}$
Hence $P (1)$ and $P (- 2)$ are $6$ and $12$ respectively.

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If p(x)=x3+3x2-2x+4 then find p(1) and p(-2)

Find the value of PGiven that P(x)=x3+3x2–2x+4P(-2)=-23+3-22–2-2+4=-8+12+4+4=12P(1)=13+312–21+4=1+3-2+4=6Hence P(1) and P-2 are 6 and 12 respectively.

If $p (x) = x^{3} + 3 x^{2} - 2 x + 4$ then find $p (1)$ and $p (- 2)$