## 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 - Page 440: 18

(5x-3)(6x-1)

#### Work Step by Step

The terms of the trinomial is of the form $ax^{2}+bx+c$ whereby a=30, b = -23 and c = 3 Find two numbers where the product is $a \times c$ or $30 \times 3 = 90$, and their sum is equal to $b$ or -23 Factors of 90: 1 $\times$ 90, Sum = 91 2 $\times$ 45, Sum = 47 3 $\times$ 30, Sum = 33 5 $\times$ 18, Sum = 23 6 $\times$ 15, Sum = 21 9 $\times$ 10, Sum = 19 From above, we see that the only numbers that fit the above condition is 5 and 18. As b is negative, but the product $a \times c$ is positive, both a and c have to be negative. Thus the two numbers are -5 and -18. Write bx as a sum of the two numbers, so -23x is written as a sum of -5x and -18x: 30$x^{2}$ -23x+3= 30$x^{2}$ -5x-18x+3 Factorize the equation by grouping: 30$x^{2}$ -5x-18x+3 = 5x(6x-1)-3(6x-1) =(5x-3)(6x-1)

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