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 12 B 2 + 22 B H − 4 H 2 as follows. Rewrite the expression as, 12 B 2 + 22 B H − 4 H 2 = 2 ( 6 B 2 + 11 B H − 2 H 2 ) (1) Consider the expression 6 B 2 + ( 11 H ) B − 2 H 2 . It is observed that, the coefficients are a = 6 , b = 11 H and c = − 2 H 2 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 6 ( − 2 H 2 ) = − 12 H 2 Thus, the value of a c is − 12 H 2 . Step 2: Find the two number, e and f, whose product is a c = − 12 H 2 and whose sum is b = 7 . Obtain the list of integer factors of − 12 H 2 such that product is − 12 H 2 and sum is 11 H as shown below in Table 1. Integers whose product is − 12 H 2 H , − 12 H − H , 12 H 6 H , − 2 H − 6 H , 2 H Sum –11 H 11 H 4 H –4 H Table 1 From Table 1, it is observed that the numbers whose product is − 12 H 2 and sum is 11 H are − H and 12 H . Let the variables e = − H and f = 12 H

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Chapter 6.2, Problem 50E

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To factor: The expression 12B2+22BH−4H2 completely.

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Chapter 6.2, Problem 49E

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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 expression 12 B 2 + 22 B H − 4 H 2 completely.

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