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(C) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$.Given $p(0) = 1$,substituting $x = 0$ gives $c = 1$.According to the Remainder Theorem,if $

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$p(-2) = 19$

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(C) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$.Given $p(0) = 1$,substituting $x = 0$ gives $c = 1$.According to the Remainder Theorem,if $p(x)$ is divided by $(x - 1)$,the remainder is $p(1) = 4$.Substituting $x = 1$ into $p(x) = ax^2 + bx + 1$,we get $a + b + 1 = 4$,which simplifies to $a + b = 3$.Similarly,if $p(x)$ is divided by $(x + 1)$,the remainder is $p(-1) = 6$.Substituting $x = -1$ into $p(x) = ax^2 + bx + 1$,we get $a - b + 1 = 6$,which simplifies to $a - b = 5$.Adding the two equations $(a + b = 3)$ and $(a - b = 5)$,we get $2a = 8$,so $a = 4$.Substituting $a = 4$ into $a + b = 3$,we get $4 + b = 3$,so $b = -1$.Thus,the polynomial is $p(x) = 4x^2 - x + 1$.To find $p(-2)$,substitute $x = -2$: $p(-2) = 4(-2)^2 - (-2) + 1 = 4(4) + 2 + 1 = 16 + 2 + 1 = 19$.

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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