Question

Which among the following is a factor of $x^2+\frac{x}{6}-\frac{1}{6}?$

• A. $3x+1$
• B. $2x+1$
• C. $x-\frac{1}{5}$
• D. $x-\frac{1}{2}$

Answer 1

The correct option is B $2x+1$

$x^2+\frac{x}{6}-\frac{1}{6}$

$=\frac{6x^2+x-1}{6}$

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

Now, we will factorise the polynomial by splitting its middle term.

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

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

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

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

Therefore, $\frac{1}{6},(3x-1)$ and $(2x+1)$ are factors of $x^2+\frac{x}{6}-\frac{1}{6}$.