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 3 x 2 + x y − 14 y 2 as follows. Consider the expression 3 x 2 + ( y ) x − 14 y 2 , where x is a variable and y is a constant. It is observed that, the coefficients are a = 3 , b = y and c = − 14 y 2 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 3 ( − 14 y 2 ) = − 42 y 2 Thus, the value of a c is − 42 y 2 . Step 2: Find the two number, m and n, whose product is a c = − 42 y 2 and whose sum is b = y . Obtain the list of integer factors of − 42 y 2 such that product is − 42 y 2 and sum is y as shown below in Table 1. Integers whose product is − 42 y 2 y , − 42 y 2 y , − 21 y − 2 y , 21 y − 6 y , 7 y 6 y , − 7 y Sum –41 y –19 y 19 y y – y Table 1 From Table 1, it is observed that the numbers whose product is − 42 y 2 and sum is y are − 6 y and 7 y

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 6.2, Problem 30E

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To factor: The given expression 3x2+xy−14y2 completely.

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Chapter 6.2, Problem 29E

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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 given expression 3 x 2 + x y − 14 y 2 completely.

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To factor: The given expression 3x2+xy−14y2 completely.

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Chapter 6 Solutions