Question

The equation having the multiple root of the equation $x^4 + 4x^3 - 16x - 16 = 0$ as its root is

• A. $x^2 + 2x - 3 = 0$
• B. $x^2 - 3x + 2 = 0$
• C. $x^2 + x - 2 = 0$
• D. $x^2 - 4x + 3 = 0$

Hint:

A key property in the theory of equations is that a multiple root of $f(x)=0$ is also a root of $f'(x)=0$. This reduces the problem of finding a multiple root of a high-degree polynomial to finding a common root between the polynomial and its derivative, which is of a lower degree and often easier to solve.

Answer 1

The Correct Option is C

Step 1: Find Multiple Root Let \( f(x) = x^4 + 4x^3 - 16x - 16 \). A multiple root is also a root of \( f'(x) = 0 \). \( f'(x) = 4x^3 + 12x^2 - 16 = 4(x^3 + 3x^2 - 4) \). By observation, \( x=1 \) makes \( 1+3-4=0 \). So \( x=1 \) is a candidate. Check \( f(1) = 1 + 4 - 16 - 16 \neq 0 \). Not the root. Factor \( f'(x) \): \( (x-1)(x^2+4x+4) = (x-1)(x+2)^2 \). Roots of \( f'(x) \) are \( 1, -2, -2 \). Check \( x=-2 \) in \( f(x) \): \( f(-2) = 16 + 4(-8) - 16(-2) - 16 = 16 - 32 + 32 - 16 = 0 \). So \( x=-2 \) is the multiple root.
Step 2: Check Options We need an equation satisfied by \( x = -2 \). (A) \( 4 - 4 - 3 \neq 0 \) (B) \( 4 + 6 + 2 \neq 0 \) (C) \( 4 - 2 - 2 = 0 \) (Satisfied) (D) \( 4 + 8 + 3 \neq 0 \)