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 27 F 3 + 21 F 2 − 48 F as follows. Rewrite the expression as, 27 F 3 + 21 F 2 − 48 F = 3 F ( 9 F 2 + 7 F − 16 ) (1) Consider the expression 9 F 2 + 7 F − 16 . It is observed that, the coefficients are a = 9 , b = 7 and c = − 16 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 9 ( − 16 ) = − 144 Thus, the value of a c is –144. Step 2: Find the two number, e and f, whose product is a c = − 144 and whose sum is b = 7 . Obtain the list of integer factors of –144 such that product is –144 and sum is 7 as shown below in Table 1. Integers whose product is –144 1 , – 144 − 1 , 144 − 9 , 16 9 , − 16 Sum –143 143 7 –7 Table 1 From Table 1, it is observed that the numbers whose product is –144 and sum is 7 are − 9 and 16 . Let, the integers take the variables e = − 9 and f = 16 . Step 3: Write a x 2 + b x + c = a x 2 + m x + n x + c

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 34RE

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To factor: The expression 27F3+21F2−48F completely.

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Chapter 6, Problem 33RE

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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 27 F 3 + 21 F 2 − 48 F completely.

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