Question

If $ \alpha, \beta, \gamma $ are the roots of the equation $ x^3 + 3x^2 + 4x + 5 = 0 $, then the cubic equation whose roots are $ 1 + 4\alpha, 1 + 4\beta, 1 + 4\gamma $ is:

• A. \( x^3 + 9x^2 - 21x + 267 = 0 \)
• B. \( x^3 + 9x^2 + 43x - 267 = 0 \)
• C. \( x^3 + 9x^2 + 41x + 267 = 0 \)
• D. \( x^3 + 9x^2 + 43x + 267 = 0 \)

Hint:

To find a new polynomial from transformed roots like \( a\alpha + b \), substitute \( x = \frac{y - b}{a} \) into the original polynomial, then simplify and clear denominators.

Answer 1

The Correct Option is D

Step 1: Let the original roots be \( \alpha, \beta, \gamma \) of \[ f(x) = x^3 + 3x^2 + 4x + 5 = 0 \]
Step 2: Define a transformation \( y = 1 + 4x \Rightarrow x = \frac{y - 1}{4} \). Substitute \( x = \frac{y - 1}{4} \) into \( f(x) \): \[ f\left(\frac{y - 1}{4}\right) = \left(\frac{y - 1}{4}\right)^3 + 3\left(\frac{y - 1}{4}\right)^2 + 4\left(\frac{y - 1}{4}\right) + 5 \] Step 3: Simplify step by step.
Let us denote: \[ u = \frac{y - 1}{4} \Rightarrow f(u) = u^3 + 3u^2 + 4u + 5 \] Now compute \( f(u) \) as a function of \( y \): \[ u = \frac{y - 1}{4} \Rightarrow f(u) = \left(\frac{y - 1}{4}\right)^3 + 3\left(\frac{y - 1}{4}\right)^2 + 4\left(\frac{y - 1}{4}\right) + 5 \] Multiply entire expression by \( 4^3 = 64 \) to eliminate denominators: \[ \Rightarrow \text{Let } g(y) = 64 \cdot f\left( \frac{y - 1}{4} \right) \] After simplification (or using a symbolic calculator): \[ g(y) = y^3 + 9y^2 + 43y + 267 \] Hence, the required cubic equation is: \[ x^3 + 9x^2 + 43x + 267 = 0 \]