## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 6 - Section 6.4 - Factoring Trinomials of the Form ax2+bx+c by Grouping - Exercise Set: 11

#### Answer

Chapter 6 - Section 6.4 - Exercise Set: 11 (Answer) Factorize : $15x^2 - 23x + 4$ a) The two numbers are -3 and -20. b) -23x to be re-written as -20x - 3x c) $15x^2 - 23x + 4$ = $(3x - 4)(5x - 1)$

#### Work Step by Step

Chapter 6 - Section 6.4 - Exercise Set: 11 (Solution) Factorize : $15x^2 - 23x + 4$ First, to look for two numbers whose product is +60 and whose sum is -23. As the two numbers have a positive product and a negative sum, pairs of negative factors of 60 are to be investigated only. Factors of 60 $\Longleftrightarrow$ Sum of Factors -1,-60 $\Longleftrightarrow$ -61 (Incorrect sum) -2,-30 $\Longleftrightarrow$ -32 (Incorrect sum) -3,-20 $\Longleftrightarrow$ -23 (Correct sum) a) The two numbers are -3 and -20. b) -23x to be re-written as -20x - 3x c) $15x^2 - 23x + 4$ = $(15x^2 - 20x - 3x + 4)$ (from the two correct numbers worked out) = $(15x^2 - 20x) - (3x - 4)$ (Group the terms) = $5x(3x - 4) - (3x - 4)$ (Factor each group) = $(3x - 4)(5x - 1)$ (Factor out (3x - 4)) Thus, $15x^2 - 23x + 4$ = $(3x - 4)(5x - 1)$

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