Factor completely: $36 x ^ { 2 } + 12 x + 1$

Question

Quizplus · Accepted Answer

This is a perfect square trinomial of the form $(ax+b)^2$, where $a = 6$ and $b = \frac{1}{6}$. We can check by expanding $(6x+\frac{1}{6})^2 = 36x^2 + 2x + \frac{1}{36}$. Notice that if we multiply this by 3, we get $36x^2+12x+\frac{1}{3}$, so we can write the original expression as $36x^2+12x+1 = 3[(6x+\frac{1}{6})^2]$. Therefore, the answer is $(6x+\frac{1}{6})^2$, or choice B in simplified form, $(6x+1)^2$.

Factor Completely 36x2+12x+136 x ^ { 2 } + 12 x + 136x2+12x+1

Factor Completely $36 x ^ { 2 } + 12 x + 1$