Factor.
-$x^{2}+\frac{2}{11} x+\frac{1}{121}$
A) Prime
B) $x\left(x+\frac{2}{11}\right)$
C) $\left(x+\frac{1}{11}\right)^{2}$
D) $\left(x+\frac{2}{11}\right)^{2}$

Question

Quizplus · Accepted Answer

The given expression is a perfect square trinomial. It can be factored as $\left(x+\frac{1}{11}\right)^{2}$ because $\left(\frac{1}{11}\right)^2 = \frac{1}{121}$ and $2 \cdot \frac{1}{11} = \frac{2}{11}$, matching the middle term.

Factor - x2+211x+1121x^{2}+\frac{2}{11} x+\frac{1}{121}x2+112​x+1211​ A) Prime B) x(x+211)x\left(x+\frac{2}{11}\right)x(x+112​) C) (x+111)2\left(x+\frac{1}{11}\right)^{2}(x+111​)2 D) (x+211)2\left(x+\frac{2}{11}\right)^{2}(x+112​)2

Factor
- $x^{2}+\frac{2}{11} x+\frac{1}{121}$ A) Prime
B) $x\left(x+\frac{2}{11}\right)$ C) $\left(x+\frac{1}{11}\right)^{2}$ D) $\left(x+\frac{2}{11}\right)^{2}$