Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 33


$(2+5i)(2-5i) = 29$

Work Step by Step

1. $(2+5i)(2-5i)$ 2. Use the difference of two squares formula $a^{2}-b^{2} = (a+b)(a-b)$ (or Multiply the two binomials together (use the FOIL method to distribute the First, Outer, Inner, and Last terms)):$4-25i^{2}$ 3. Change $-25i^{2}$ to 25, because $i^{2} = -1$ and $-25\times-1 = 25$:$4+25$ 4. Add the two numbers together:$29$
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