What are Integer Factor Pairs?
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An integer factor pair is a pair of integers that are factors of another number.\n\nFor example:\n5 has one integer factor pair: 1 and 5.\n10 has two integer factor pairs: 1 and 10, and 2 and 5.\n100 has five integer factor pairs: 1 & 100, 2 & 50, 4 & 25, 5 & 20, and 10 & 10.
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Prime factor each\n4 = 2 x 2\n7 = 7\n9 = 3 x 3\n\nTake union of all factors, removing duplicates (none here)\nAnswer = 2 x 2 x 3 x 3 x 7 = 252
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2, 2, 2, 3, 3. The prime factors of a number are the set of prime numbers which, when all multiplied together, make that number.\n\nYou can use this program to find prime factors:\nhttp://www.brennen.net/primes/FactorApplet.html
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(x+3)(x^2-x+4)/(x+3) common factor\n\nmultiply by x+3\n\nx^2-x+4
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factors of 27 are 1 and 27, 3 and 9\nfactors of 54 are 1 and 54, 2 and 27, 6 and 9
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You look for factors of 4 that add to -5\nFactors of 4:\n1,4\n2,2\n-1,-4\n-2,-2\n\n-1 and -4 add to -5\nAnswer: (x-1)(x-4)
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1. 2 or 0\nGCF= greatest common factor?\n2. 12x\n3. 2x\n\nnotation for power is 2^5
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Instead of just giving you the answer as I did for your other question, let me explain what GCF and LCM are, and give you a general method for finding them.\n\nThe GCF, or greatest common factor of two numbers, is the largest integer which is a factor of both numbers - that is, both numbers can be divided by it with no remainder. There are three general methods for finding the GCF:\n\n1: guess and check\n\nTo find the GCF of two numbers A and B, start with the lower of the two numbers (let's assume it's B), and check all the numbers B, B/2, B/3, B/4... 1 to determine whether they are integers, and if they are, whether they are factors of A. The first number tested that meets both of these criteria is the GCF. This method is the simplest, but also by far the slowest, unless you get lucky and the GCF is one of the first few numbers tested.\n\n2. prime factor decomposition\n\nDecompose both A and B into products of prime factors (this step is not trivial). Then find all of the prime factors that appear in BOTH the prime decompositions of A and B, and if any factors appear multiple times in either product, write down that factor the same number of times that it appears in the product that contains the least number of that factor. Then find the product of these numbers to get the GCF of A and B. For instance, if your numbers are 728 and 196, your prime decompositions would be (2*2*2*7*13) and (2*2*7*7), thus your GCF would be (2*2*7), which is 28. The most time-consuming step in this method is finding the prime factors, however, many of the numbers you work with in math class and in real life have obvious prime factorizations, so when working with small numbers (2-3 digits), this is often the best method in practice.\n\n3: Euclid's method\n\nFor finding the GCF of large numbers, neither of the above methods are usually sufficient. Fortunately, there is a faster method. Let A mod B be the remainder of A divided by B - e.g. 9 mod 4=1. Then we may find the GCF of A and B as follows: find A mod B. If the answer is zero, then B is the GCF and we are done. Otherwise, find the GCF of B and A mod B. Repeat this procedure until you get zero. The last number you divided by is your GCF.\n\nExample: finding the GCF of 78920 and 65109.\n78920 mod 65109=13811\n65109 mod 13811=9865\n13811 mod 9865=3946\n9865 mod 3946=1973\n3946 mod 1973=0\n\nThus the GCF of 78920 and 65109 is 1973.\n\nWhy this works: the Euclidian algorithm is based on the observation that if two numbers have a common factor, then the remainder of these numbers after division also has that factor. This is easily shown as follows: if q is the quotient of A/B and r is the remainder, then A = qB + r, with q and r integers. Divide both sides by f: A/f = qB/f + r/f and thus r/f = A/f - qB/f. Suppose f is a factor of both A and B: then A/f and B/f are also integers, and as q is an integer, qB/f is also an integer. Since the difference of two integers is also an integer, r/f is an integer and thus f is a factor of r. Q.E.D.\n\nThe LCM, or least common multiple, is the smallest integer that both numbers are a factor of - that is, the LCM can be divided by either of them with no remainder. Finding the LCM is pretty simple once you know how to find the greatest common factor: simply divide the product of A and B by the greatest common factor of A and B. That is: LCM(A, B) = A*B/GCF(A, B).
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the factors of 27 are 3-3-3.  when you have two of the same number in a factor tree, you can take them outside the root.  So now you have 3squareroot3.  squareroot3 + 3squareroot3 = 4 squareroot 3.  Make sense?
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You are looking for factor of -5 such that when you add them together you get -4\nFactors of -5"\n-5 and 1 or 5 and -1\n-5+1=-4 and 5-1=4\nThus, we are looking for the -5 and 1\nx^2 - 4x - 5 = 0\n(x-5)(x+1) = 0\nwhich implies x-5 = 0 or x+1 = 0 which gives you\nx = 5 or x = -1\nCheck your answer by plugging these values back into the original equation.
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x=[2+/-rt(4+60)]/2\n=[2+/-8]/2\n=-3 or 5\nso the factors are\n(x+3)(x-5)
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One of the factors will be x-x=0.\nSo, the answer is 0.\nThis is because 0 times any number is 0.
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x^n + y^n has factors when n is not a power of 2: n = 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, ...\n\nIf n is odd then ( x + y ) is a factor, by the factor theorem: put y = -x.\n\nIf n has an odd factor n = p q where p is odd, then x^(pq) = (x^q)^p. So, x^n + y^n = (x^q)^p + (y^q)^p can be factored. Example: x^44 + y^44 = (x^4)^11 + (y^4)^11 has x^4 + y^4 as a factor.\n\nThat only leaves x^(2^n) + y^(2^n): x^2 + y^2, x^4 + y^4, x^8 + y^8, ... None of these can be factored, using Eisenstein's criteria.\n\nIn the above I assume that all coefficients are integers. If not, note that\n\nx^4 + y^4 = ( x² + √2 x + 1 ) ( x² - √2 x + 1 )\n\nand similar results for other powers of 2.
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I am giving you a method which would minimize number of possibilities.\nFactorize the number of factors( in your example 9) in as many way as possible.( in your example 9 and 3*3)\nNow find out the possible numbers ( one for one factor combination) in following way ( we are going to have 2 numbers)\nThen all factor combinations are sorted so that largest one is in first place and lowest one in last place.\nFor a factor combination x1*x2*x3*...xn*...\nthe number will be \n2^(x1-1) * 3^(x2-1) * 5^(x3-1) * ... * pn(xn-1)...\nwhere pn is nth prime number.\nFor our two combinations,\nfor 9-> 2^(9-1)=256\nfor 3*3-> 2^(3-1) * 3^(3-1) = 36\nLowest of these numbers is our result. \nIn our example it is 36.\nLet us see another example lowest number with 12 factors.\n12 factorizes to 12, 4*3, 6*2, 3*2*2\nfor 12-> 2^(12-1)=2048\nfor 4*3-> 2^(4-1) * 3^(3-1) = 72\nfor 6*2-> 2^(6-1) * 3^(2-1) = 96\nfor 3*2*2-> 2^(3-1) * 3^(2-1) * 5^(2-1)=60.\nSo our answer is 60.\nThis method will be if the number of factor cobination  is too large. An way to get good answer( may not be correct) is to start from prime factorization and continue untill collecting first two terms into one causes better result.\n\nThat is in case of 12\nWe will start from 2*2*3 and get 60.\nThen test if 4*3 is better. If not we will take 60 as answer.( as it is 72 so we will).\nThis case we have correct result but modified version may not produce correct result.
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If the zeros are: -4, 0, and 2, our factored equation must look like this:\nf(x) = x(x+4)(x-2)\n\nFoil it out:\n\nf(x) = x(x2 + 4x - 2x - 8)\nf(x) = x(x2 + 2x -8)\n\nf(x) = x3 + 2x2 -8x\n\nAnswer number 3.\n\nHope that helps.
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why don't you get a piece of paper and make a factor tree?\nlearn something yourself.
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factor 9x^2y^2-9y^2\n9y^2(x^2-1)\n9y^2(x+1)(x-1)