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 − 5 x − 2 as follows. It is observed that, the coefficients are a = 3 , b = − 5 and c = − 2 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 3 ( − 2 ) = − 6 Thus, the value of a c is − 6 . Step 2: Find the two number, m and n, whose product is a c = − 6 and whose sum is b = − 5 . Obtain the list of integer factors of − 6 such that product is − 6 and sum is − 5 as shown below in Table 1. Integers whose product is − 6 1 , − 6 2 , − 3 − 2 , 3 − 1 , 6 Sum –5 –1 1 5 Table 1 From Table 1, it is observed that the numbers whose product is − 6 and sum is − 5 are 1 and − 6

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

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ISBN: 9780137554843

Author: Allyn Washington, Richard Evans

Publisher: PEARSON+

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Chapter 6.2, Problem 19E

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To factor: The given expression 3x2−5x−2 completely.

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Chapter 6.2, Problem 18E

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Chapter 6 Solutions

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

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 − 5 x − 2 completely.

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