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 4 p 2 − 25 p q + 6 q 2 as follows. Rewrite the expression as, 4 p 2 − 25 p q + 6 q 2 = 4 p 2 − ( 25 q ) p + 6 q 2 . It is observed that, the coefficients are a = 4 , b = − 25 q and c = 6 q 2 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 4 ( 6 q 2 ) = 24 q 2 Thus, the value of a c is 24 q 2 . Step 2: Find the two number, e and f, whose product is a c = 24 q 2 and whose sum is b = − 25 q . Obtain the list of integer factors of 24 q 2 such that the product is 24 q 2 and the sum is − 25 q as shown below in Table 1. Integers whose product is 24 q 2 q , 24 q − q , − 24 q 12 q , 2 q − 2 q , − 12 q Sum 25 q − 25 q 14 q − 14 q Table 1 From Table 1, it is observed that the numbers whose product is 24 q 2 and sum is − 25 q are − q and − 24 q

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 43E

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To factor: The expression 4p2−25pq+6q2 completely.

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Chapter 6.2, Problem 42E

Chapter 6.2, Problem 44E

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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 4 p 2 − 25 p q + 6 q 2 completely.

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To factor: The expression 4p2−25pq+6q2 completely.

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Chapter 6 Solutions