Answer
Chapter 6 - Section 6.2 - Exercise Set: 85 (Answer)
$x^2 + \frac{1}{2}x + \frac{1}{16}$ = $(x + \frac{1}{4})^2$
Work Step by Step
Chapter 6 - Section 6.2 - Exercise Set: 85 (Solution)
Factor : $x^2 + \frac{1}{2}x + \frac{1}{16}$
First, take out the GCF of $\frac{1}{16}$ from the trinomial
$x^2 + \frac{1}{2}x + \frac{1}{16}$ = $\frac{1}{16}(16x^2 + 8x + 1)$
Let $(16x^2 + 8x + 1)$ = $(\triangle x + 1)(\square x + 1)$
Next, to look for two numbers whose product is 16 and whose sum is 8.
Factors of 16 $\Longleftrightarrow$ Sum of Factors
1,16 $\Longleftrightarrow$ 17 (Incorrect sum)
2,8 $\Longleftrightarrow$ 10 (Incorrect sum)
4,4 $\Longleftrightarrow$ 8 (Correct sum, so the two numbers are 4 and 4)
Thus, $(16x^2 + 8x + 1)$ = $(4x + 1)(4x + 1)$ = $(4x + 1)^2$
And, $x^2 + \frac{1}{2}x + \frac{1}{16}$
= $\frac{1}{16}(4x + 1)^2$
= $[\frac{1}{4}(4x + 1)]^2$
= $(x + \frac{1}{4})^2$
