Question

$\left\{x\in R:\frac{2x-1}{x^3+4x^2+3x}\right\}$ is equal to:

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

Answer 1

The correct option is C

$R-\left\{0,-1,-3\right\}$

Explanation for the correct option

The expression is $\frac{2x-1}{x^3+4x^2+3x}$

For the expression to be real, the denominator should not be equal to zero. Factorize the denominator to find the values of $x$that make the denominator equal to zero.

$\Rightarrow$ $x^3+4x^2+3x=0$

$\Rightarrow$ $x\left(x^2+4x+3\right)=0$

$\Rightarrow$ $x\left(x^2+3x+x+3\right)=0$

$\Rightarrow$$x\left\{x\left(x+3\right)+1\left(x+3\right)\right\}=0$

$\Rightarrow$ $x\left(x+1\right)\left(x+3\right)=0\left[\because factorizebysplittingthemiddleterm\right]$

$\Rightarrow$ $x=0,-1,-3$

The value of $x$can be all real numbers except $0,-1,-3$ i.e., $x\in \left[R-\left\{0,-1,-3\right\}\right]$

Hence, the correct option is C.