Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 8 - Graphing Quadratic Functions - 8.5 - Using Intercept Form - Exercises - Page 456: 44

Answer

$9$ does not belong to the solution set

Work Step by Step

The mistake is in the second zero, which is not $9$. $$y=(x+4)(x^2-9)$$ Factor the function completely: $$y=(x-(-4))(x-3)(x+3)$$ $$y=(x-(-4))(x-3)(x-(-3))$$ Now it is in the form $y=a(x-p)(x-q)(x-z)$, so the zeros are $-4$ ,$3$ and $-3$. The function has three zeros and $9$ is not among them.
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