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 - Page 429: 49

Answer

Chapter 6 - Section 6.2 - Exercise Set: 49 (Answer) $4x^2y + 4xy - 12y$ = $4y(x^2 + x - 3)$

Work Step by Step

Chapter 6 - Section 6.2 - Exercise Set: 49 (Solution) Factorize : $4x^2y + 4xy - 12y$ First step : Take out the GCF of $4x^2y$, $4xy$ and $12y$ which is $4y$ $4x^2y + 4xy – 12y$ = $4y(x^2 + x - 3)$ Take $(x^2 + x - 3)$ to be $(x + \triangle)(x + \square)$ For this, we have to look for two numbers whose product is -3 and whose sum is +1. Factors of -3 $\Longleftrightarrow$ Sum of Factors 1,-3 $\Longleftrightarrow$ -2 (Incorrect sum) -1,3 $\Longleftrightarrow$ 2 (Incorrect sum) As there are no two numbers that can have a product of -3 and a sum of +1, no further factorization can be done with $(x^2 + x - 3)$. Thus, $4x^2y + 4xy - 12y$ = $4y(x^2 + x - 3)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.