Algebra: A Combined Approach (4th Edition)

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

Chapter 6 - Section 6.4 - Factoring Trinomials of the Form ax2+bx+c by Grouping - Exercise Set: 12

Answer

Chapter 6 - Section 6.4 - Exercise Set: 12 (Answer) Factorize : $6x^2 - 13x + 5$ a) The two numbers are -3 and -10. b) -13x to be re-written as -3x - 10x c) $6x^2 - 13x + 5$ = $(2x - 1)(3x - 5)$

Work Step by Step

Chapter 6 - Section 6.4 - Exercise Set: 12 (Solution) Factorize : $6x^2 - 13x + 5$ First, to look for two numbers whose product is +30 and whose sum is -13. As the two numbers have a positive product and a negative sum, pairs of negative factors of 30 are to be investigated only. Factors of 30 $\Longleftrightarrow$ Sum of Factors -1,-30 $\Longleftrightarrow$ -31 (Incorrect sum) -2,-15 $\Longleftrightarrow$ -17 (Incorrect sum) -3,-10 $\Longleftrightarrow$ -13 (Correct sum) a) The two numbers are -3 and -10. b) -13x to be re-written as -3x - 10x c) $6x^2 - 13x + 5$ = $(6x^2 - 3x - 10x + 5)$ (from the two correct numbers worked out) = $(6x^2 - 3x) - (10x - 5)$ (Group the terms) = $3x(2x - 1) - 5(2x - 1)$ (Factor each group) = $(2x - 1)(3x - 5)$ (Factor out (2x - 1)) Thus, $6x^2 - 13x + 5$ = $(2x - 1)(3x - 5)$
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