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.5 - Page 74: 77


$x^2+64$ $=x^2+8^2$ $=(x+8i)(x-8i)$

Work Step by Step

To factorise this expression we must first find the square root of the constant. In this case the constant is 64 and the square root of 64 is 8. The expression is now written in the form: $A^2+B^2$. Now that the square root of the constant is found the expression can be factorised using the difference of two squares rule for a positive constant: $A^2+B^2=(A+Bi)(A-Bi)$. In one set of brackets the constant ($B$) is positive and in the other it is negative. The $i$ is an imaginary number an can be thought of as $\sqrt -1$.
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