Question:
Let \(p(x)\) be a quadratic polynomial such that \(p(0) = 1\). If \(p(x)\) leaves remainder 4 when divided by \(x - 1\) and it leaves remainder 6 when divided by \(x + 1\); then :
Let \(p(x)\) be a quadratic polynomial such that \(p(0) = 1\). If \(p(x)\) leaves remainder 4 when divided by \(x - 1\) and it leaves remainder 6 when divided by \(x + 1\); then :
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Use the remainder theorem: \(p(1) = 4\), \(p(-1) = 6\), and \(p(0) = 1\) gives the constant term.
Updated On: Nov 11, 2026
- \(p(2) = 11\)
- \(p(2) = 19\)
- \(p(-2) = 19\)
- \(p(-2) = 11\)
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The Correct Option is C
Solution and Explanation
Let \(p(x) = ax^2 + bx + c\). \(p(0) = 1\) gives \(c = 1\).
Remainder theorem: \(p(1) = a + b + 1 = 4\), so \(a + b = 3\).
\(p(-1) = a - b + 1 = 6\), so \(a - b = 5\).
Hence \(a = 4\), \(b = -1\) and \(p(x) = 4x^2 - x + 1\).
\(p(2) = 16 - 2 + 1 = 15\) and \(p(-2) = 16 + 2 + 1 = 19\).
So option (3) \(p(-2) = 19\) is correct.
Answer: (3) \(p(-2) = 19\)
Remainder theorem: \(p(1) = a + b + 1 = 4\), so \(a + b = 3\).
\(p(-1) = a - b + 1 = 6\), so \(a - b = 5\).
Hence \(a = 4\), \(b = -1\) and \(p(x) = 4x^2 - x + 1\).
\(p(2) = 16 - 2 + 1 = 15\) and \(p(-2) = 16 + 2 + 1 = 19\).
So option (3) \(p(-2) = 19\) is correct.
Answer: (3) \(p(-2) = 19\)
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