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(D) Since $p(x)$ has relative extrema at $x=1$ and $x=2$,we have $p'(x) = 0$ at $x=1$ and $x=2$.Thus,$p'(x) = A(x-1)(x-2) = A(x^2 - 3x + 2)$.Integrati

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$-12$

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(D) Since $p(x)$ has relative extrema at $x=1$ and $x=2$,we have $p'(x) = 0$ at $x=1$ and $x=2$.Thus,$p'(x) = A(x-1)(x-2) = A(x^2 - 3x + 2)$.Integrating $p'(x)$,we get $p(x) = A(\frac{x^3}{3} - \frac{3x^2}{2} + 2x) + C$.Given $p(1) = 8$,we have $8 = A(\frac{1}{3} - \frac{3}{2} + 2) + C = A(\frac{2-9+12}{6}) + C = \frac{5A}{6} + C$,so $48 = 5A + 6C$ (Equation $1$).Given $p(2) = 4$,we have $4 = A(\frac{8}{3} - 6 + 4) + C = A(\frac{8-6}{3}) + C = \frac{2A}{3} + C$,so $12 = 2A + 3C$ (Equation $2$).Multiplying Equation $2$ by $2$,we get $24 = 4A + 6C$.Subtracting this from Equation $1$,we get $48 - 24 = (5A - 4A) + (6C - 6C)$,which gives $A = 24$.Substituting $A = 24$ into Equation $2$,we get $12 = 2(24) + 3C$,so $12 = 48 + 3C$,which implies $3C = -36$,so $C = -12$.Since $p(0) = C$,we have $p(0) = -12$.

If $p(x)$ is a polynomial of degree three that has a local maximum value $8$ at $x=1$ and a local minimum value $4$ at $x=2$,then $p(0)$ is equal to:

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