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 6 t 4 + t 2 − 12 as follows. Rewrite the expression 6 t 4 + t 2 − 12 as, 6 t 4 + t 2 − 12 = 6 ( t 2 ) 2 + ( t 2 ) − 12 . It is observed that, the coefficients are a = 6 , b = 1 and c = − 12 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 6 ( − 12 ) = − 72 Thus, the value of a c is –72. Step 2: Find the two number, e and f, whose product is a c = − 72 and whose sum is b = 1 . Obtain the list of integer factors of –72 such that product is –72 and sum is 1 as shown below in Table 1. Integers whose product is –72 1 , − 72 72 , − 1 9 , − 8 − 9 , 8 Sum –71 71 1 –1 Table 1 From Table 1, it is observed that the numbers whose product is –72 and sum is 1 are − 8 and 9 . Let, the variables e = − 8 and f = 9 . 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 6.2, Problem 40E

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To factor: The given expression 6t4+t2−12 completely.

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Chapter 6.2, Problem 39E

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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 6 t 4 + t 2 − 12 completely.

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To factor: The given expression 6t4+t2−12 completely.

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Chapter 6 Solutions