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).

Answer 2

Given:
The cubic equation is:
\[ ax^3 + bx^2 + cx + 1 = 0 \]
It is given that \( -1 \) is a twice repeated root. That means:
The roots are: \( -1, -1, r \) (for some real root \( r \))

Step 1: Use factorized form of the cubic
Since \( -1 \) is a repeated root, we can write the polynomial as:
\[ a(x + 1)^2(x - r) \]
Now expand this expression:
First, expand \( (x + 1)^2 = x^2 + 2x + 1 \)
Then:
\[ a(x^2 + 2x + 1)(x - r) \]
Multiply:
\[ a[(x^3 - rx^2 + 2x^2 - 2rx + x - r)] = a(x^3 + (2 - r)x^2 + (1 - 2r)x - r) \]

Step 2: Compare with standard form
Standard form is: \( ax^3 + bx^2 + cx + 1 \)
Compare with expanded form:
\[ ax^3 + a(2 - r)x^2 + a(1 - 2r)x - ar \]
From this, we match coefficients:
- \( b = a(2 - r) \)
- \( c = a(1 - 2r) \)
- Constant term: \( -ar = 1 \Rightarrow r = -\frac{1}{a} \)

Step 3: Substitute value of r
We now substitute \( r = -\frac{1}{a} \) into expressions for \( b \) and \( c \):

For \( b \):
\[ b = a\left(2 - \left(-\frac{1}{a}\right)\right) = a\left(2 + \frac{1}{a}\right) = 2a + 1 \]

For \( c \):
\[ c = a\left(1 - 2\left(-\frac{1}{a}\right)\right) = a\left(1 + \frac{2}{a}\right) = a + 2 \]

Final Answer:
The values are: \( b = 2a + 1, \quad c = a + 2 \)