Question

Factors of $(x^2+\frac{x}{6}-\frac{1}{6})$ are

• A. $\frac{1}{6}(2x+1)(3x+1)$
• B. $\frac{1}{6}(2x+1)(3x-1)$
• C. $\frac{1}{6}(2x-1)(3x-1)$
• D. $\frac{1}{6}(2x-1)(3x+1)$

Answer 1

Answer: (B)

$\frac{1}{6}(2x+1)(3x-1)$

$(x^2+\frac{x}{6}-\frac{1}{6})=\frac{1}{6}[6x^2+x-1]$

$\Rightarrow \frac{1}{6}[6x^2+3x-2x-1]$

$\Rightarrow \frac{1}{6}\left[3x\right(2x+1)-(2x+1\left)\right]$

$\Rightarrow \frac{1}{6}(2x+1)(3x-1)$