Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises - Page 33: 8

Answer

(a) No (b) Yes (c) Yes (d) Yes

Work Step by Step

(a) Using the perfect square factoring formula $(A + B)^{2}$ = $A^{2}$ + 2AB + $B^{2}$, we can conclude that $(x + 5)^{2}$ = $x^{2}$ + 10x + 25, rather than $x^{2}$ + 25. (b) Using the perfect square factoring formula $(A + B)^{2}$ = $A^{2}$ + 2AB + $B^{2}$, we can conclude that $(x + a)^{2}$,, where a $\ne$ 0, indeed yields three terms when expanded (namely, those would be $x^{2}$, 2xa, and $a^{2}$. (c) Using the difference of squares factoring formula $A^{2}$ - $B^{2}$ = (A - B)(A + B), we can see that the expression (x + 5)(x - 5) is indeed equal to $x^{2}$ - 25, in which (x + 5) and (x - 5) correspond to (A + B) and (A - B). (d) Again, using the difference of squares factoring formula $A^{2}$ - $B^{2}$ = (A - B)(A + B), we see that (x + a)(x - a), given that a $\ne$ 0, would indeed yield two terms, namely, $x^{2}$ - $a^{2}$.
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