Answer
$21x^4-41x^2+10$
Work Step by Step
$$(7x^2-2)(3x^2-5)$$ $$=(7x^2\times 3x^2)+(7x^2\times (-5))+(-2\times 3x^2)+(-2\times (-5))$$ $$=21x^4-35x^2-6x^2+10$$ $$=21x^4-41x^2+10$$
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.4 - Page 61: 28
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