Explanation Procedure used: Factoring general trinomials by grouping : In a trinomial of the form a x 2 + b x + c , a ≠ 1 , split up the x -term into two terms using the method described below. “Step 1: Find the product a c . Step 2: Find two numbers whose product is a c and whose sums is b . Step 3: Replace the middle term bx with two x -terms having these numbers as coefficients. That is, a x 2 + b x + c = a x 2 + m x + n x + c . Step 4: Complete the factorization by grouping”. Calculation: Factor the given expression 10 b 2 + 23 b − 5 as follows. It is observed that, the coefficients are a = 10 , b = 23 and c = − 5 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 10 ( − 5 ) = − 50 Thus, the value of a c is –50. Step 2: Find the two number, e and f, whose product is a c = − 50 and whose sum is b = 23 . Obtain the list of integer factors of –50 such that product is –50 and sum is 23 as shown below in Table 1. Integers whose product is –50 1 , − 50 − 1 , 50 − 2 , 25 2 , − 25 Sum –49 49 23 –23 Table 1 From Table 1, it is observed that the numbers whose product is –50 and sum is 23 are − 2 and 2 5 . Let, the integers take the variables e = − 2 and f = 25

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 6, Problem 35RE

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To factor: The expression 10b2+23b−5 completely.

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Chapter 6, Problem 34RE

Chapter 6, Problem 36RE

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Chapter 6 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

Ch. 6.1 - Prob. 1PE Ch. 6.1 - Prob. 2PE Ch. 6.1 - Prob. 3PE Ch. 6.1 - Prob. 4PE Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E

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To factor: The expression 10 b 2 + 23 b − 5 completely.

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To factor: The expression 10b2+23b−5 completely.

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Chapter 6 Solutions