What's the easiest way to learn factoring polynomials?
129715
Be familiar with the divisors of small numbers.  Learn to quickly look at two numbers and know if one is the sum or difference of two numbers that multiply to the other.\n\nBe able to do arithmetic in your head.\n\nLearn how to do synthetic division.\n\nPractice on many examples.
621189
5) find the factors\nsquareroot of x * x * x * x * x * x /  7*7*5*5*2*2\nthe factors can come out of the square root by pairs only\nyou have 3 pairs of x's = x^3\nyou have 1 pair of 7's so it comes out of the square root as a 7\nx^3/7*5*2\nx^3/70 is it's simplified term\n\n__________________\nif u want all the correct answers then that would tarenirator's answers
21064
The instruction is to write a polynomial function with the zeros 0, 1, and 1+i. Please keep in mind that when a zero consists radicals or imaginary numbers, the conjugate of that zero is also a zero. For exmaple, here 1+i is a zero. Then 1-i must also be a zero.\n\nThis problem is like working backwards. Instead of factoring the polynomial, you are asked to do it backwards, starting with the zeros. Since 0, 1, and 1+i and 1-i are zeros, you would expect the last step of the factoring to look something like this:\n\nx(x-1)[x-(1+i)][x-(1-i)]\n\nNow simplify this expression a bit. It becomes:\n\nx^2-x[(x-1-i)(x-1+i)]\n\nSimplify this expression even further and you get:\n\nx^4-3x^3+4x^2-2x
386466
The easiest function of x are:\n\nf(x) = constant\nf(x) = a*x, where a is a constant : 1st order poly\nf(x) = b*x^2 where b is a constant 2nd order\nf(x) = c*x^3 where c is a constant: 3rd order\n...\nf(x) = z*x^n, where z is a constant, n is a positive integer, nth order polynomial\n\na general polynomial is a sum of these.  They are convenient to work with for lots of reasons.  That's why they are so important.
611709
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!
295841
x^3 + 4 = 3x^2\n\nx^3 - 3x^2 + 4 = 0\n\nx^3 + 1 - 3(x^2 - 1) = 0\n\n(factorising)\n\n(x+1)(x^2 - x + 1) - 3(x+1)(x-1) = 0\n\n[x+1](x^2 - 4x + 4) = 0\n\n(x+1)(x-2)(x-2) = 0\n\nx= 2, -1
223739
Yes there are.\nThere is answers.yahoo.com, http://mathforum.org/dr.math/ if you are feeling dumb, http://www.webmath.com/ and more.\n\nBTW, oh pls, you are telling me simple algebraic equations take forever!? I was doing them when i was 11 (quick and easy)
130462
You need to factor everything, then cancel common factors.\n4x+8 factors as 4(x+2)\n6x-3x^2 factors as 3(2-x)\n3x^2-4x-4 factors as (3x+2)(x-2)\n9x^2-4 factors as (3x+2)(3x-2)\n\nCancel the (3x+2)'s, and the (x-2) can cancel with the (2-x) leaving a factor of -1.  The factors that are left are:\n[-4(x+2)]/[3(3x-2)].
130686
Well. Do you know how to work rational expressions in general? The first link below is a useful general resource and has a section on rational expressions and polynomial long division; the second is specifically on rational expressions at college intro-level.\n\nI used to tutor math and I found that most people's difficulty with word problems was not the actual math, it was extracting the appropriate equations from the words. Once they had the equations, they could solve them (assuming they'd learned the basics). \n\nI'd find it pretty difficult to give you general guidelines on "how to solve word problems with rational expressions" but the main trick with word problems is to figure out a) what variables are involved, and assign each of them a letter (e.g., speed of first train is x, time it has been travelling is y, speed of second train is z...), b) sort out the relationship between them (e.g. x = 2z/5), and c) make sure you know what *exactly* is being asked. It is not at all uncommon for word problems to include extraneous information (colour of train, age of driver) to muddle the issue.\n\nYour best bet is almost certainly to go ask your teacher, professor, TA, tutor, study group, or classmate-who-understands-it for help. They will be better able to figure out where your problem is if they can watch you work through an actual example.
307256
1.)\nstep one: if they are zeroes, use opp. \nexample....\n (x-2) (x+9) (x-7)          if you solve for x it should be opposite of given zeroes.\n\nstep two: FOIL the first two...\nyou'll get   x^ +9x -2x -18 which simplifies to\n\n(x^2 + 7x -18) (x-7) \n\nstep three: FOIL these two together...\nyou'll get     x^3 +7x^2 - 18x -7x^2 -49x + 126\n\nwhich simplifies to  \n   x^3 - 67x +126\n\n\n2.) I don't know about the second one... I could solve it if I spent some time, but start with...\n (x - 4i) (x+5) (x + 2i)\nthan just do what I did in the problem above, factor the first two, than factor your answer with the last one... and remember that i^2 is equal to 1...
210139
Actually, -18 *does* have something to do with it. Try to find two numbers which multiply to give -18 and which add to give -5. In this case, it doesn't work, so the polynomial doesn't factor.\n\nConsider 8x^2-10x-3. Multiply the 8 and -3 to get\n-24. Now find to numbers which multiply to give -24 and which add to give -10. In this case -12 and 2 work. Now do this:\n\n8x^2-10x-3=\n8x^2-12x+2x-3= (using the -12+2=-10)\n4x(2x-3)+(2x-3)= (factoring 1st two and last two terms)\n(4x+1)(2x-3).\n\nThis gives a better way to factor that simple guessing. It will always work if the quadratic can be factored.
363762
Try using a search engine, such as google, and type in "Algebra Help". Here are some sites I found;\n\nThese sites relate to FACTORING\n- http://www.purplemath.com/modules/factnumb.htm\n- http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm\n\nThis site has general algebra information, like printable worksheets, instruction on doing various problems, and calculators.\n- http://www.algebrahelp.com/
541488
I got   3(x² - 5x + 4) or 3( x - 4 )( x - 1 )\n\nLet me show you my work...\n\n3( x - 2 )² - 3(x - 2) - 6\n\nSimplify\n\n3( x -2 )( x - 2 ) - 3( x- 2 ) - 6\n\nDistribute\n\n3( x² - 4x + 4) - 3x + 6 - 6\n\nDistribute and Simplify\n\n3x² - 12x + 12 - 3x\n\nSimplify\n\n3x² - 15x + 12\n\nYou can now simplify by factoring out 3\n\n3(x² - 5x + 4)\nor\n3( x - 4 )( x - 1 )
476367
Including negative factors:\n-8, -4, -2, -1, 1, 2, 4, 8\n\nJust positive factors:\n1, 2, 4, 8\n\nPrime factorization:\n2 x 2 x 2 = 8
138785
It's pol_Y_nomial (polynomial) and try this one:\n\nhttp://www.google.com/u/lamarmathnotes?q=polynomials\n\nor this one:\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
70773
alright, you need to understand this much more than you need the correct answer. You won't have Yahoo! answers to help you on any tests.\n\nYou need help factoring. A real-life tutor who can talk to you and show you in person is best, but I'm going to try to help you understand. \n\na polynomial is in the form ax^2 + bx + c. This is from the formula of (Z*x + q) * (R*x + p). Don't let all these letters scare you. They are just place holders for numbers. \n\nThe only thing that can get the letter A in Ax^2 +bx +c is the first two parts of the "terms" you need to multiply together. With letters a= z*r. In order to get C, you only need to worry about the last parts in the terms. Using letters c= q*p. B is what makes sure you know what two terms were multiplied together. B = z*p + q*r. In order to "factor" out a term from a polynomial you have to find the set of z, p, q, and r that works. I'll show you now with numbers.\n\nLet's say you have the equation 2x^2 + 8x +8. See, it's in the form of ax^2 +bx + c. a=2, b=8, c=8. NOw you have to figure out the facotrs of 2 and 8 (which are a and c)\nfor a=2 the numbers that are multiplied together to get two(the factors) are 2 and 1. Since you know that only the first two parts of the things you multiply together have to get you a, you can set up something that looks like this: \n(2x + ?) *(x + ?) \n\nNow you need to find what two numbers fit into your equation to get you c=8. The factors of 8 are 1*8 or 2*4. Let's try putting 1*8 into the equation.\n(2x+1)* (x+8) = 2x^2 +17x + 8. That's not right, so lets try to reverse them.\n(2x+8)* (x+1) = 2x^2 +10x + 8. That's not right either, so we know only the factors of 2*4 will work in this equation.\nLet's try them\n\n(2x+2)*(x+4) = 2x^2 +10x +8. It may seem like we've tried each of the factors, but we can still reverse 2*4.\n\n(2x+4) * (x+2) = 2x^2 +8x+8. That's right. Once you get more practice and understand this better, you can "see" certain factors that have to be included just because "factoring" will be automatic. But, just to make sure you should always check you answer like I did when I showed you the example.\n\nNow that you hopefully understand the hard part, perfect squares are easy. All you do is take  the square root of 'a' and take the square root of 'c' and put them into the equation. When it is a PERFECT SQUARE this works because of: ax^2 + bx +c = (z*x+ q)^2. ax^2 = (zx)*(zx) so a = z^2, and the square root of a = z. c = (q)*(q) so c = q^2, and the square root of c= q. I'll use numbers now.\n\n100x^2 + 240x + 144. a = 100, so the quare root of 100 = 10. z=10. Then the square root of 144 is 12, so q=12. That means y=(10x+12)^2. Now, to find b, all you have to do is foil it out.
345973
1.\n3x + 6/((x + 1)(x + 2)) Factor 3 out of the top.\n[3(x + 2)]/[(x + 1)(x + 2)] x+2's cancel\n3/(x + 1) <-- This is simplified.\n\nI don't have the energy to do the other ones right now.  :/