Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.6 Perform Operations with Complex Numbers - 4.6 Exercises - Mixed Review - Page 282: 85


The solutions are $-3$ and $4$.

Work Step by Step

$ 3x^{2}-3x-36=0\qquad$ ...factor out $3$ from left side. $ 3(x^{2}-x-12)=0\qquad$ ...divide entire expression with $3$. $ x^{2}-x-12=0\qquad$ ...rewrite the expression as $(kx+m)(lx+n)$ where $k$ and $l$ are two numbers such that their sum is $-1$ and product is $12$. $\qquad$ ...those numbers are $-4$ and $3$. $\qquad$ ...$-4+3=-1,\ -4\cdot 3=-12$ $\qquad$ ...therefore, $x^{2}-x-12=(x-4)(x+3)$ $\qquad$ ...apply the zero product property and solve for $x$. $x-4=0$ or $x+3=0$ $x=4$ or $x=-3$ The solutions are $-3$ and $4$.
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