If the remainders of the polynomial f(x) when | vedclass

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(A) Let the remainder be $Q(x) = ax^2 + bx + c$. Since the divisor is a cubic polynomial,the remainder must be at most a quadratic polynomial.By the R

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$2x^2 - 3x + 1$

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(A) Let the remainder be $Q(x) = ax^2 + bx + c$. Since the divisor is a cubic polynomial,the remainder must be at most a quadratic polynomial.By the Remainder Theorem,we have:$f(-1) = 6 \implies Q(-1) = a - b + c = 6$ $(i)$$f(2) = 3 \implies Q(2) = 4a + 2b + c = 3$ (ii)$f(-2) = 15 \implies Q(-2) = 4a - 2b + c = 15$ (iii)Subtracting (iii) from (ii): $(4a + 2b + c) - (4a - 2b + c) = 3 - 15 \implies 4b = -12 \implies b = -3$.Substituting $b = -3$ into (ii) and (iii):$4a - 6 + c = 3 \implies 4a + c = 9$ (iv)$4a + 6 + c = 15 \implies 4a + c = 9$ (consistent).Substituting $b = -3$ into $(i)$: $a - (-3) + c = 6 \implies a + c = 3$ $(v)$.Subtracting $(v)$ from (iv): $(4a + c) - (a + c) = 9 - 3 \implies 3a = 6 \implies a = 2$.Substituting $a = 2$ into $(v)$: $2 + c = 3 \implies c = 1$.Thus,the remainder is $Q(x) = 2x^2 - 3x + 1$.

If the remainders of the polynomial $f(x)$ when divided by $(x + 1), (x - 2), (x + 2)$ are $6, 3, 15$ respectively,then the remainder of $f(x)$ when divided by $(x + 1)(x + 2)(x - 2)$ is:

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