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 e 2 + 18 e − 1560 as follows. Rewrite the expression 3 e 2 + 18 e − 1560 as, 3 e 2 + 18 e − 1560 = 3 ( e 2 + 6 e − 520 ) (1) Consider the expression e 2 + 6 e − 520 . It is observed that, the coefficients are a = 1 , b = 6 and c = − 520 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 1 ( − 520 ) = − 520 Thus, the value of a c is –520. Step 2: Find the two number, m and n, whose product is a c = − 520 and whose sum is b = 6 . Obtain the list of integer factors of –520 such that product is –520 and sum is 6 as shown below in Table 1. Integers whose product is –520 1 , − 520 − 26 , 20 26 , − 20 − 1 , 520 Sum –519 –6 6 519 Table 1 From the Table 1, it is observed that the numbers whose product is –520 and sum is 6 are − 20 and 26 . Let the variables m = − 20 and n = 26 . Step 3: Write a x 2 + b x + c = a x 2 + m x + n x + c

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Chapter 6.2, Problem 58E

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To factor: The expression 3e2+18e−1560 completely.

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Chapter 6.2, Problem 57E

Chapter 6.2, Problem 59E

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Chapter 6 Solutions

EBK BASIC TECHNICAL MATHEMATICS

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 3 e 2 + 18 e − 1560 completely.

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To factor: The expression 3e2+18e−1560 completely.

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