Add the proper constant to the binomial so that the resulting trinomial is a perfect square trinomial. Then factor the
trinomial.
-$$x ^ { 2 } + \frac { 1 } { 6 } x$$
A) $x ^ { 2 } + \frac { 1 } { 6 } x + 144 = ( x + 12 ) ^ { 2 }$
B) $x ^ { 2 } + \frac { 1 } { 6 } x + \frac { 1 } { 144 } = \left( x + \frac { 1 } { 12 } \right) ^ { 2 }$
C) $x ^ { 2 } + \frac { 1 } { 6 } x + \frac { 1 } { 36 } = \left( x + \frac { 1 } { 6 } \right) ^ { 2 }$
D) $x ^ { 2 } + \frac { 1 } { 6 } x + \frac { 1 } { 12 } = \left( x + \frac { 1 } { 6 } \right) ^ { 2 }$

Question

Quizplus · Accepted Answer

To form a perfect square trinomial from $x^2 + \frac{1}{6}x$, we need to add $\left(\frac{1}{2} \cdot \frac{1}{6}\right)^2 = \left(\frac{1}{12}\right)^2 = \frac{1}{144}$. Thus, the trinomial becomes $x^2 + \frac{1}{6}x + \frac{1}{144} = \left(x + \frac{1}{12}\right)^2$, which matches option B.

Add the Proper Constant to the Binomial So That the Resulting