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Divide the sum of the three consecutive odd integers by 3: 402/3 = 134. The smallest of these integers will be two less than 134 and the largest will be two more than 134, so the three consecutive odd integers will be 132, 134, and 136.

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โˆ™ 2009-02-22 12:41:49
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Algebra

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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: Find three consecutive integers with sum 402?
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Yes: 398, 399, 400, 401 and 402.


What times what equals 402?

2 x 3 x 67 = 402 2 x 201 = 402 3 x 134 = 402 6 x 67 = 402 To solve such a problem you use prime decomposition: This allows you to break any integer down into a multiple of prime numbers as in 2x3x67=402. Two, Three and 67 are all prime. From there to find all possible options, multiply the primes together: 2x3x67=402 --> 3x67=201 --> 2x201=402 2x3x67=402 --> 2x67=134 --> 3x134=402 2x3x67=402 --> 2x3=6 --> 6x67=402 This is under the assumption that this problem is restricted to integers. If real numbers are used, there are infinitely many solutions:


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