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 10 R 4 − 6 R 2 − 4 as follows. Rewrite the expression 10 R 4 − 6 R 2 − 4 as, 10 R 4 − 6 R 2 − 4 = 2 [ 5 ( R 2 ) 2 − 3 ( R 2 ) − 2 ] (1) Consider the expression 5 ( R 2 ) 2 − 3 ( R 2 ) − 2 . It is observed that, the coefficients are a = 5 , b = − 3 and c = − 2 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 5 ( − 2 ) = − 10 Thus, the value of a c is –10. Step 2: Find the two number, m and n, whose product is a c = − 10 and whose sum is b = − 3 . Obtain the list of integer factors of –10 such that product is –10 and sum is –3 as shown below in Table 1. Integers whose product is –10 1 , − 10 5 , − 2 − 5 , 2 − 1 , 10 Sum –9 3 –3 9 Table 1 From Table 1, it is observed that the numbers whose product is –10 and sum is –3 are − 5 and 2 . Let, the variables m = − 5 and n = 2 . Step 3: Write a x 2 + b x + c = a x 2 + m x + n x + c . Rewrite the expression 5 ( R 2 ) 2 − 3 ( R 2 ) − 2 in the form a R 2 + m R + n R + c as, 5 ( R 2 ) 2 − 3 ( R 2 ) − 2 = 5 ( R 2 ) 2 + 2 ( R 2 ) − 5 ( R 2 ) − 2 Step 4: Factor the expression in step 3 by grouping

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

11th Edition

ISBN: 9780137554843

Author: Allyn Washington, Richard Evans

Publisher: PEARSON+

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Chapter 6.2, Problem 26E

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To factor: The given expression 10R4−6R2−4 completely.

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Chapter 6.2, Problem 25E

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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 given expression 10 R 4 − 6 R 2 − 4 completely.

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