Answer:
Yes, it is possible to get the same remainder when the divisor is different. This is known as the Remainder Theorem, which states that if you divide a polynomial \( f(x) \) by \( x - a \), the remainder is \( f(a) \).
Let's find the remainder when \( f(a) = 2a^4 + 5a^3 - 14a^2 - 5a + 15 \) is divided by \( a - 1 \) and \( a + 1 \):
1. Remainder when divided by \( a - 1 \):
\[ f(1) = 2(1)^4 + 5(1)^3 - 14(1)^2 - 5(1) + 15 \]
\[ f(1) = 2 + 5 - 14 - 5 + 15 \]
\[ f(1) = 3 \]
2. Remainder when divided by \( a + 1 \):
\[ f(-1) = 2(-1)^4 + 5(-1)^3 - 14(-1)^2 - 5(-1) + 15 \]
\[ f(-1) = 2 - 5 - 14 + 5 + 15 \]
\[ f(-1) = 3 \]
So, the remainder is the same (3) when \( f(a) \) is divided by both \( a - 1 \) and \( a + 1 \).
Step-by-step explanation:
Mark me as brainleist
