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 a 2 c 2 − 6 a c + 3 as follows. Rewrite the expression 3 a 2 c 2 − 6 a c + 3 as, 3 a 2 c 2 − 6 a c + 3 = 3 ( a 2 c 2 − 2 a c + 1 ) (1) Consider the expression ( a c ) 2 − 2 ( a c ) + 1 . It is observed that, the coefficients are a = 1 , b = − 2 and c = 1 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 1 ( 1 ) = 1 Thus, the value of a c is 1. Step 2: Find the two number, m and n, whose product is a c = 1 and whose sum is b = − 2 . Obtain the list of integer factors of 36 such that product is 1 and sum is –2 as shown below in Table 1. Integers whose product is 1 1 , 1 − 1 , − 1 Sum 2 –2 Table 1 From Table 1, it is observed that the numbers whose product is 1 and sum is –2 are − 1 and − 1 . Let, the variables m = − 1 and n = − 1 . Step 3: Write a x 2 + b x + c = a x 2 + m x + n x + c

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 6.2, Problem 38E

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To factor: The given expression 3a2c2−6ac+3 completely.

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Chapter 6.2, Problem 37E

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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 a 2 c 2 − 6 a c + 3 completely.

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To factor: The given expression 3a2c2−6ac+3 completely.

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