Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.2 - Factoring Trinomials of the Form x2+bx+c - Exercise Set: 67

Answer

Chapter 6 - Section 6.2 - Exercise Set: 67 (Answer) $\frac{1}{2}y^2 - \frac{9}{2}y - 11$ = $\frac{1}{2}(y + 2)(y - 11)$

Work Step by Step

Chapter 6 - Section 6.2 - Exercise Set: 67 (Solution) Factor : $\frac{1}{2}y^2 - \frac{9}{2}y - 11$ First, take out the GCF of $\frac{1}{2}$ from the polynomial $\frac{1}{2}y^2 - \frac{9}{2}y - 11$ = $\frac{1}{2}(y^2 - 9y - 22)$ Let $(y^2 - 9y - 22)$ = $(y + \triangle)(y + \square)$ Next, to look for two numbers whose product is -22 and whose sum is -9. Factors of -22 $\Longleftrightarrow$ Sum of Factors 1,-22 $\Longleftrightarrow$ -21 (Incorrect sum) 2,-11 $\Longleftrightarrow$ -9 (Correct sum, so the two numbers are 2 and -11) 11,-2 $\Longleftrightarrow$ 9 (Incorrect sum) 22,-1 $\Longleftrightarrow$ 21 (Incorrect sum) Thus, $(y^2 - 9y – 22)$ = $(y + 2)(y - 11)$ And, $\frac{1}{2}y^2 - \frac{9}{2}y - 11$ = $\frac{1}{2}(y + 2)(y - 11)$
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