Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises: 67

Answer

$\color{blue}{(p^2+25)(p+5)(p-5)}$

Work Step by Step

Note that $p^4=(p^2)^2$ and $625=25^2$. Thus, the given binomial is equivalent to: $=(p^2)^2-25^2$ Factor the binomial using the formula $a^2-b^2=(a+b)(a-b)$ where $a=p^2$ and $b=25$ to obtain: $=(p^2+25)(p^2-25) \\=(p^2+25)(p^2-5^2)$ Use the same formula above to factor the second binomial with $a=p$ and $b=5$ to obtain: $\\=\color{blue}{(p^2+25)(p+5)(p-5)}$Note that $p^4=(p^2)^2$ and $625=25^2$. Thus, the given binomial is equivalent to: $=(p^2)^2-25^2$ Factor the binomial using the formula $a^2-b^2=(a+b)(a-b)$ where $a=p^2$ and $b=25$ to obtain: $=(p^2+25)(p^2-25) \\=(p^2+25)(p^2-5^2)$ Use the same formula above to factor the second binomial with $a=p$ and $b=5$ to obtain: $\\=\color{blue}{(p^2+25)(p+5)(p-5)}$
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