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 35: 133

Answer

$(a+b+c)(a- b+c)(a+b-c)(-a+ b+c)$

Work Step by Step

Let $E=4a^{2}c^{2} - (a^{2} - b^{2} + c^{2})^{2}$ First we factor it out as the difference of squares: $E=(2ac-(a^{2} - b^{2} + c^{2}))(2ac+(a^{2} - b^{2} + c^{2}))$ we can take out the parenthesis of both terms by multiplying by $-1$ on left term and by $1$ on right term: $E=(2ac- a^{2} + b^{2} - c^{2})(2ac+a^{2} - b^{2} + c^{2})$ Let's take the left term and later we will come to the right one left term: $2ac- a^{2} + b^{2} - c^{2}$ we rearrange it to be $- a^{2}+2ac - c^{2}+ b^{2}$ = $-( a^{2}-2ac + c^{2}- b^{2})$ = $ b^2-(a- c)^{2}$ = $b-a+c)(b+a-c)$ = $(-a+ b+c)(a+b-c)$ right term: $2ac+a^{2} - b^{2} + c^{2}$ we rearrange it to be $a^{2} + 2ac + c^{2} - b^{2}$ =$(a+c)^{2} - b^{2}$ =$(a+ c- b)(a+c+b)$ =$(a- b+c)(a+b+c)$ then by combining both terms it would be: $E=(a+b+c)(a- b+c)(a+b-c)(-a+ b+c)$
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