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(A) Let the remainder be $R(x) = ax^2 + bx + c$ since the divisor is a cubic polynomial.By the Remainder Theorem,we have:$f(-1) = 6 \implies a(-1)^2 +

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$2x^2 - 3x + 1$

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(A) Let the remainder be $R(x) = ax^2 + bx + c$ since the divisor is a cubic polynomial.By the Remainder Theorem,we have:$f(-1) = 6 \implies a(-1)^2 + b(-1) + c = 6 \implies a - b + c = 6$ $(i)$$f(2) = 3 \implies a(2)^2 + b(2) + c = 3 \implies 4a + 2b + c = 3$ (ii)$f(-2) = 15 \implies a(-2)^2 + b(-2) + c = 15 \implies 4a - 2b + c = 15$ (iii)Subtracting (iii) from (ii): $(4a + 2b + c) - (4a - 2b + c) = 3 - 15 \implies 4b = -12 \implies b = -3$.Substituting $b = -3$ into $(i)$: $a - (-3) + c = 6 \implies a + c = 3 \implies c = 3 - a$.Substituting $b = -3$ and $c = 3 - a$ into (ii): $4a + 2(-3) + (3 - a) = 3 \implies 3a - 3 = 3 \implies 3a = 6 \implies a = 2$.Then $c = 3 - 2 = 1$.Thus,the remainder is $R(x) = 2x^2 - 3x + 1$.

If a polynomial $f(x)$ is divided by $(x + 1)$,$(x - 2)$,and $(x + 2)$,the remainders are $6$,$3$,and $15$ respectively. Find the remainder when $f(x)$ is divided by $(x + 1)(x - 2)(x + 2)$.

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