Question:
If \(x^2 + px + 1\) is a factor of \(ax^3 + bx + c\), then:
If \(x^2 + px + 1\) is a factor of \(ax^3 + bx + c\), then:
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When a quadratic is a factor of a cubic, assume the quotient is linear, expand, and equate coefficients to find relations among coefficients.
Updated On: Jul 18, 2026
- \(a^2 + c^2 = ab + 3\)
- \(a^2 - c^2 = ab\)
- \(a^2 - c^2 = -ab\)
- \(a^2 + c^2 = ab\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the factor theorem.
If \(x^2 + px + 1\) is a factor of \(ax^3 + bx + c\), then there exist a polynomial \(Q(x)\) such that: \[ ax^3 + bx + c = (x^2 + px + 1)Q(x) \]
Step 2: Assume \(Q(x) = kx + l\).
Multiply and equate coefficients: \[ (x^2 + px + 1)(kx + l) = kx^3 + (kp+l)x^2 + (lp+k)x + l \]
Step 3: Compare coefficients with \(ax^3 + 0\cdot x^2 + bx + c\).
\[ \begin{cases} a = k \\ 0 = kp + l \\ b = lp + k \\ c = l \end{cases} \]
Step 4: Solve for \(l\).
From \(c = l\) and \(0 = kp + l \Rightarrow 0 = ap + c\) \(\Rightarrow p = -c/a\)
Step 5: Solve for relation among \(a, b, c\).
\[ b = lp + k = c(-c/a) + a = a - c^2/a = (a^2 - c^2)/a \] Multiply both sides by \(a\): \[ a^2 - c^2 = ab \]
Step 6: Final conclusion.
\[ \boxed{a^2 - c^2 = ab} \]
If \(x^2 + px + 1\) is a factor of \(ax^3 + bx + c\), then there exist a polynomial \(Q(x)\) such that: \[ ax^3 + bx + c = (x^2 + px + 1)Q(x) \]
Step 2: Assume \(Q(x) = kx + l\).
Multiply and equate coefficients: \[ (x^2 + px + 1)(kx + l) = kx^3 + (kp+l)x^2 + (lp+k)x + l \]
Step 3: Compare coefficients with \(ax^3 + 0\cdot x^2 + bx + c\).
\[ \begin{cases} a = k \\ 0 = kp + l \\ b = lp + k \\ c = l \end{cases} \]
Step 4: Solve for \(l\).
From \(c = l\) and \(0 = kp + l \Rightarrow 0 = ap + c\) \(\Rightarrow p = -c/a\)
Step 5: Solve for relation among \(a, b, c\).
\[ b = lp + k = c(-c/a) + a = a - c^2/a = (a^2 - c^2)/a \] Multiply both sides by \(a\): \[ a^2 - c^2 = ab \]
Step 6: Final conclusion.
\[ \boxed{a^2 - c^2 = ab} \]
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