Question

If \( -1 \) is a twice repeated root of the equation \( ax^3 + bx^2 + cx + 1 = 0 \), then \( ax^3 + bx^2 + cx + 1 = 0 \) represents:

• A. \( b = 2a + 1, \quad c = a + 1 \)
• B. \( b = 2a + 1, \quad c = a - 2 \)
• C. \( b = 2a + 1, \quad c = a + 2 \)
• D. \( b = 2a, \quad c = a + 2 \)

Hint:

When given a repeated root, use factoring and expand to compare coefficients for solving the equation.

Answer 1

The Correct Option is C

We are given that \( -1 \) is a twice repeated root of the cubic equation \( ax^3 + bx^2 + cx + 1 = 0 \). This means the equation can be factored as: \[ a(x + 1)^2(x + r) = 0 \] where \( r \) is another root of the equation. Expand \( (x + 1)^2 \): \[ (x + 1)^2 = x^2 + 2x + 1 \] Now multiply by \( (x + r) \): \[ (x + 1)^2(x + r) = (x^2 + 2x + 1)(x + r) = x^3 + rx^2 + 2x^2 + 2rx + x + r \] \[ = x^3 + (r + 2)x^2 + (2r + 1)x + r \] Thus, the cubic equation becomes: \[ a(x^3 + (r + 2)x^2 + (2r + 1)x + r) = 0 \] Comparing the coefficients with the original equation \( ax^3 + bx^2 + cx + 1 = 0 \), we get the following system of equations: - \( b = a(r + 2) \) - \( c = a(2r + 1) \) - Constant term: \( 1 = a \cdot r \) From \( 1 = a \cdot r \), we find \( r = \frac{1}{a} \). Substituting this into the equations for \( b \) and \( c \): - \( b = a \left( \frac{1}{a} + 2 \right) = 2a + 1 \) - \( c = a \left( 2 \cdot \frac{1}{a} + 1 \right) = a + 2 \) Thus, the correct answer is option (3).