## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson

# Appendix A - Review - A.4 Factoring Polynomials - A.4 Assess Your Understanding - Page A40: 73

#### Answer

$\frac{1}{16}$, $(x-(1/4))^2$

#### Work Step by Step

To complete the square of $x^2 -\frac{1}{2}x$ we add $(b/2)^2$. In this case, the "$b$" is the second coefficient, $-1/2$, so we add $((-1/2)/2)^2=1/16$ and we have $x^2 -\frac{1}{2}x+\frac{1}{16}$. Then by factorization, we get $$x^2 -\frac{1}{2}x+\frac{1}{16}=(x-(1/4))^2 .$$

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