Question

$x \in R : \frac{2x -1}{x^3 + 4x^2 + 3x} \in R$ Equals

• A. R - {0}
• B. R- {0,1,3}
• C. R-{0, -1,-3}
• D. R- 0,-1,-3,+ $\frac{1}{2}$

Answer 1

The Correct Option is C

Let $ A = x \in R : \frac{2x -1}{x^3 + 4x^2 + 3x}$ Now , $x^3 + 4x^2 + 3x = x(x^2 + 4x + 3)$ $ = x(x+ 3) (x + 1)$ $A = R - \{0, -1, -3\}$

Answer 2

To determine the values of \( x \) for which the expression \( \frac{2x - 1}{x^3 + 4x^2 + 3x} \) belongs to the set of real numbers \( \mathbb{R} \), we need to find the values of \( x \) that make the denominator \( x^3 + 4x^2 + 3x \) non-zero.

The denominator can be factored as \( x(x^2 + 4x + 3) \), which further simplifies to \( x(x+1)(x+3) \).

For the expression to be real, the denominator must not be zero. So, the values of \( x \) that make the expression real are all real numbers except for \( x = 0, -1, -3 \).

Therefore, the set of values for \( x \) that satisfy the condition is \( \mathbb{R} - \{0, -1, -3\} \), which means all real numbers except 0, -1, and -3.