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

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. 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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Problem 11 (a) A tank is discharging water through an orifice at a depth of T meter below the surface of the water whose area is A m². The following are the values of a for the corresponding values of A: A 1.257 1.390 x 1.50 1.65 1.520 1.650 1.809 1.962 2.123 2.295 2.462|2.650 1.80 1.95 2.10 2.25 2.40 2.55 2.70 2.85 Using the formula -3.0 (0.018)T = dx. calculate T, the time in seconds for the level of the water to drop from 3.0 m to 1.5 m above the orifice. (b) The velocity of a train which starts from rest is given by the fol- lowing table, the time being reckoned in minutes from the start and the speed in km/hour: | † (minutes) |2|4 6 8 10 12 14 16 18 20 v (km/hr) 16 28.8 40 46.4 51.2 32.0 17.6 8 3.2 0 Estimate approximately the total distance ran in 20 minutes.

- Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p − 1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., p-1 2 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). 23 32 how come? The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. The set T is the subset of these residues exceeding So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1.

Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p-1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., 2 p-1 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. 23 The set T is the subset of these residues exceeding 2° So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1. how come?

Chapter 6 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

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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