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(C) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$.Given $p(0) = 1$,we have $c = 1$.By the Remainder Theorem,$p(1) = 4$ and $p(-1) = 6$.Substi

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$p(-2) = 19$

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(C) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$.Given $p(0) = 1$,we have $c = 1$.By the Remainder Theorem,$p(1) = 4$ and $p(-1) = 6$.Substituting these into $p(x) = ax^2 + bx + 1$:$p(1) = a(1)^2 + b(1) + 1 = 4 \Rightarrow a + b = 3$.$p(-1) = a(-1)^2 + b(-1) + 1 = 6 \Rightarrow a - b = 5$.Adding the two equations: $2a = 8 \Rightarrow a = 4$.Subtracting the two equations: $2b = -2 \Rightarrow b = -1$.Thus,$p(x) = 4x^2 - x + 1$.Now,evaluating $p(-2)$:$p(-2) = 4(-2)^2 - (-2) + 1 = 4(4) + 2 + 1 = 16 + 3 = 19$.Therefore,$p(-2) = 19$ is the correct statement.

Let $p(x)$ be a quadratic polynomial such that $p(0) = 1$. If $p(x)$ leaves a remainder of $4$ when divided by $x - 1$ and a remainder of $6$ when divided by $x + 1$,then:

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