College Algebra (6th Edition)

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: 45

Answer

$x^2-6x+9$

Work Step by Step

$$(x-3)^2$$ $$=(x-3)(x-3)$$ $$=(x\times x)+(x\times (-3))+(-3\times x)+(-3\times (-3))$$ $$=x^2-3x-3x+9$$ $$=x^2-6x+9$$
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