## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises: 125

#### Answer

$(a+1)(a-1)(a+2)(a-2)$

#### Work Step by Step

$Factor$ $the$ $expression$ $completely:$ $(a^2+1)^2 - 7(a^2+1) + 10$ We can identify that the expression is in trinomial form, so we can factor it into two binomials $(a^2+1)^2 - 7(a^2+1) + 10$ = $((a^2+1)-2)((a^2+1)-5)$ Simplify the binomials $= (a^2+1-2)(a^2+1-5)$ $= (a^2-1)(a^2-4)$ Use the Difference of Squares Formula for both binomials Difference of Squares Formula: $(A^2-B^2) = (A-B)(A+B)$ $(a^2-1) = (a+1)(a-1)$ $(a^2-4) = (a+2)(a-2)$ $= (a+1)(a-1)(a+2)(a-2)$

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