To find the constants \( p \) and \( q \) such that the function
\[ f(x) = 2x^3 + px^2 + qx - 5 \]
gives a remainder of 8 when divided by \( x + 1 \), and a remainder of 35 when divided by \( x + 2 \), we will use the Remainder Theorem. According to the Remainder Theorem, the remainder of the division of a polynomial \( f(x) \) by \( x - c \) is \( f(c) \).
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Finding \( p \) and \( q \):
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For the first case, \( f(-1) = 8 \): \[ f(-1) = 2(-1)^3 + p(-1)^2 + q(-1) - 5 \] Simplifying: \[ f(-1) = 2(-1) + p(1) - q - 5 = -2 + p - q - 5 = p - q - 7 \] Setting this equal to 8 gives us: \[ p - q - 7 = 8 \implies p - q = 15 \quad \text{(1)} \]
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For the second case, \( f(-2) = 35 \): \[ f(-2) = 2(-2)^3 + p(-2)^2 + q(-2) - 5 \] Simplifying: \[ f(-2) = 2(-8) + p(4) - 2q - 5 = -16 + 4p - 2q - 5 = 4p - 2q - 21 \] Setting this equal to 35 gives us: \[ 4p - 2q - 21 = 35 \implies 4p - 2q = 56 \quad \text{(2)} \]
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Solving the equations:
From (1), we have \( p - q = 15 \). Rearranging this gives: \[ p = q + 15 \]
Substituting \( p \) in equation (2): \[ 4(q + 15) - 2q = 56 \] Expanding and simplifying: \[ 4q + 60 - 2q = 56 \implies 2q + 60 = 56 \implies 2q = -4 \implies q = -2 \]
Substituting \( q \) back to find \( p \): \[ p = -2 + 15 = 13 \]
Thus we find: \[ p = 13, \quad q = -2 \]
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Finding the remainder when \( f(x) \) is divided by \( x - 2 \):
Now we can compute \( f(2) \): \[ f(2) = 2(2)^3 + 13(2)^2 - 2(2) - 5 \] Calculating: \[ = 2(8) + 13(4) - 4 - 5 = 16 + 52 - 4 - 5 \] Combining these: \[ = 16 + 52 - 4 - 5 = 16 + 52 - 9 = 59 \]
So, the remainder when \( f(x) \) is divided by \( x - 2 \) is
\[ \boxed{59}. \]