Add the proper constant to the binomial so that the resulting trinomial is a perfect square trinomial. Then factor the
trinomial.
-$$x ^ { 2 } - \frac { 2 } { 9 } x$$
A) $x ^ { 2 } - \frac { 2 } { 9 } x + \frac { 2 } { 81 } = \left( x - \frac { 1 } { 9 } \right) ^ { 2 }$
B) $x ^ { 2 } - \frac { 2 } { 9 } x + \frac { 1 } { 81 } = \left( x - \frac { 1 } { 9 } \right) ^ { 2 }$
C) $x ^ { 2 } - \frac { 2 } { 9 } x + \frac { 1 } { 81 } = \left( x + \frac { 1 } { 9 } \right) ^ { 2 }$
D) $x ^ { 2 } - \frac { 2 } { 9 } x + \frac { 4 } { 81 } = \left( x - \frac { 2 } { 9 } \right) ^ { 2 }$

Question

Quizplus · Accepted Answer

To form a perfect square trinomial from $x^2 - \frac{2}{9}x$, we need to add $\left(\frac{b}{2a}\right)^2$, where $a = 1$ and $b = -\frac{2}{9}$. Calculating gives $\left(\frac{-\frac{2}{9}}{2}\right)^2 = \left(\frac{-1}{9}\right)^2 = \frac{1}{81}$. Thus, the trinomial becomes $x^2 - \frac{2}{9}x + \frac{1}{81}$, which factors to $\left(x - \frac{1}{9}\right)^2$.

Add the Proper Constant to the Binomial So That the Resulting