Iwould like to know how to start the first step in algebra?
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Algebra help website is your basic starting point.
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since 2 is a root you take (x-2)*(x-2) and -3 is one also so you multiply (x+3) now you know it is a fifth degree so you need two more so you can take (4x^2+1) and multiply\n(x-2)(x-2)(x+3)(4X^2+1) you do the algebra
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Just do a search for "algebra." You'll find a lot of web sites, including:\nhttp://mathworld.wolfram.com/topics/Algebra.html\nhttp://www.purplemath.com/\nhttp://www.algebrahelp.com/
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Not just algebra...but anything in this world looks hard until we understand/learn it. So dont worry. \n\nNow, to give you some help regarding algebra. I I have answered a similar question before. Somebody asked what's the best way to teach algebra. I have answered them with a helpful link (see the link below), where they have video lessons of algebra for free. A professor teaches you algebra from scratch and gives real-world examples from time to time. He makes it look so easy that you not only would find algebra easy but interesting too. Here is the link.\n\nhttp://64.78.42.182/free-ed/Math/Algebra/Algebra01_VidLect.asp
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Look up "Algebra Solved 2005" or "algebrator", these two programs not only solve problems, but they give step by step solutions for understanding. they have tutorials and practice test. I went from knowing none to fairly advanced in a few weeks.
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Here you go!\n\nhttp://www.math.com/homeworkhelp/Algebra.html
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Wow, there is a lot here!  I'll start with some tips for solving algebra problems.  The first thing you should always remember is to play the equal game.  Whenever you add, subtract, multiply, or divide, you should do the same to the other side.  The other tip is to use the order of operations when simplifying.  Remember the sentence: Please excuse my dear Aunt Sally.  The initial letters of each word stands for the order of operations.  Do the Parenthesis first, followed by the Exponents, then Multiplication or Division, and Addition and/or subtraction next.  Finally, you should always simplify first using the order of operations before you start shifting numbers from one side of the = to the other.  \n\nI'm not going to work all of these problems, but I'll show you a few to get you started.\n\n1)\n3x-7 = 2   Add 7 to both sides\n3x = 9   Divide by 3\nx = 3\n\n4)\n6x - 3 = 2x - (4-3)/3   Simplify the stuff inside the parenthesis first\n6x - 3 = 2x - 1/3  Add 3 to both sides\n6x = 2x - 5/3  Subtract 2x from both sides\n4x = -5/3  Multiply both sides by 3\n12x = -5  Divide both sides by 12\nx = -12/5\n\nI hope this helps.  Good luck!
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use arithmatic progression formula n(2a+(n-1)d)/2\nwhere n=no. of terms. a=value of first term and d=common difference between each consecutive term\nsay fro your first eg. no. of black ball = 1+3+5+...\nn=10 a=1 d=2\nso sum = 10(2+9*2)/2 = 100\nsolve the rest by changing values of n, a and other variables respectively
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The key to math is understanding the definitions. Make sure you understand them first and then start on examples of problems.  You need to ask yourself two questions when approaching math problems.  1. What is the questions asking me? 2. What does the answer look like?  For example:\n\nFind the roots for the function y=x^2 + 7x + 12\n\nYou know that your answer needs to be where y=0.  Now you need to know what the answer looks like. It can take varying forms but these would be reasonable:\n\n(-3,0) and (-4,0) or x=-3 and x=-4\n\nRemember to always understand what is being asked of you and to have an idea of what it should look like.  The biggest problem students have is relying on the calculator.  The calculator can only tell you what you tell it.  So if you don't understand what it should look like then it may come out wrong.
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go online, click on search and type in MATH 012 : Intermediate algerbra course. something should pop up. and then write down the evidence
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It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.\n\nAddressing your first question, let's write it out like a long division problem.  You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:\n\nx^2 + 3x - 2  )  x^4 + x^3 - 5x^2 + 13x - 6\n\nOur first step is to look at the first term in each polynomial, and divide the smaller into the larger.  x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2.  You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.\n\nNow, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2.  Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2).  Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2.  We then drop down the next term from the dividend, the 13x term, and repeat:\n\n-2x^3 / x^2 = -2x (add that term to our quotient)\n\n-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x\n\n(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x\n\nFinally, we drop down the last term from the dividend, -6, and repeat once more.  I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!\n\n3x^2 / x^2 = 3 (so we add a +3 term to our quotient)\n\n3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6\n\n(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!\n\nSo our final quotient is x^2 - 2x + 3, with no remainder.\n===\nNow, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5.  We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.\n\nSorry there's no better way to write this out on Yahoo Answers.  Hope that helped anyway!
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Use the following four steps to solve a quadratic equation, in this order:\n\nStep 1: Guess the answer\n\nIf x^2=0, can you guess the answer?\nIf x^2=1, you can guess, but remember that there may be two answers. (1 and -1)\n\nYou can tell if you are guessing right by plugging back into the original equation. If it worked, you're done.\n\nIf you can't guess in 30 seconds, you are done with Step 1. Go to Step 2.\n\n\nStep 2: Try factoring\n\nx^2 + 4x + 4 = (x+2)(x+2)\n\nRemember that the constant term of your factors has to multiply to the constant term of the quadratic. \n\nFor example, if you have x squared plus something plus 6, remember that your constant terms must be multiples of 6. \n\nTry:\n\n(x + 1)(x + 6)\n(x - 1)(x - 6)\n(x + 2)(x + 3)\n(x - 2)(x - 3)\n\nYou just ran out of factors of 6. If this didn't work, go to step 3.\n\n\nStep 3 -- Complete the Square\n\nOne important thing to remember about completing the square is that it does exactly the same thing as the quadratic equation. Pick either Step 3 or Step 4 as your last step. Guessing and factoring don't always work because you can't always guess and you can't always factor. But completing the square and using the quadratic formula will always find the answer if there is one.\n\nSuppose you have \n\nx^2 - 6x - 16 = 0\n\nHere's how to complete the square:\n\n#1. Move the constant (whatever it is) to the right side\n\nx^2 - 6x = 16\n\n#2. Find half of the coeficient of the x term and add the square of that number to both sides of the equation:\n\nThe x term here is -6x\nThe coefficient is -6\nHalf of -6 is -3 (Remember this number -3)\n-3 sqared is 9. Add 9 to both sides of the equation:\n\nx^2 - 6x + 9 = 16 + 9\n\n#3. If you did #2 correctly, then the left side is a perfect square and the constant term is -3 (remember -3?)\n\n(x - 3)^2 = 25\n\n(If you're not sure, multiply it out and see if you get what you had before.)\n\n#4. Take the square root of both sides. Remember, it could be negative or positive.\n\nx - 3 = 5\nx - 3 = -5\n\n#5. Solve your two equations\n\nx = 8\nx = -2\n\n#6. Put your answers back into the original equation and see if you got the right answer. \n\n\nStep 4. \n\nIf you did Step 3, you already have the answer, and you're already done. If you didn't do Step 3, use the quadratic formula to get the answer. You can find the quadratic formula here:\n\nhttp://en.wikipedia.org/wiki/Quadratic_formula\n\nMemorize the quadratic formula and never forget it.  Write it on your forhead until you have it memorized.
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Until you change your mind about math, you're not going to be able to help, no matter how much you want to.  Algebra isn't bizarre.  It's actually a pretty basic way to find an unknown. And you use it every day without realizing it.   For example, let's say you have $1.00.  You go to the store to buy apples.  They each cost 20 cents.  How many could you buy for $1.00?  That's algebra in its most basic form.  \n\nSolving algebra problems is no different than solving any other math problem.  Begin by breaking it down into what you know and what you need to know.  And always remember that whatever you do to one side of the equation must be done to the other side. \n\nTo help with your son's work, do it together.  He needs to learn the process more than he needs to get the right answer.  Work problems together and talk them out.  You'll learn from one another, and you'll really cherish the time together.
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You have to do the equations in the [ ] first.\nThen you have to do multiplication before addition.\nHere's the order of how you should do math like this:\nP - Parenthesis\nE - Exponents (a number to the something power)\nM - Multiplication\nD - Division\nA - Addition\nS - Subtraction\n\nSo you do [2+6(-3)] first\n[2-18]\n[-16]\nNow it's 15-3[-16]\n15+48\nAnswer: 63
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"The Complete Idiot's Guide to Algebra" \n\nRotten title - great resource
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There's lots of help online. Try these web sites:\n\nhttp://www.algebrahelp.com/\nhttp://www.purplemath.com/\nhttp://www.sosmath.com/algebra/algebra.html\nhttp://www.helpalgebra.com/\n\nThere's even more. So start with these, and good luck!
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This acronym helped me out a lot when trying to figure out algebra. PEMDAS (Please Excuse My Dear Aunt Sally) It shows you the order in which you solve the problem. First off you start with anything in parentheses. Then you solve the Exponents. Next comes Multiplying or Dividing. And finally you Add or Subtract. Hope this helps!