can someone please refer me to a website to explain polonomial functions?
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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
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www.bbc.co.uk/revisewise\nwww.murderousmaths.co.uk\nwww.a729.piczo.com\nwww.bbc.co.uk/bitesize
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math.com is an awsome site to go to, i use it for my hw help!
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If you really meant "slops" follow this link:\n\n\nhttp://mathforum.org/library/drmath/view/55824.html
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First are the answers, and their individual in-depth explanations are below.  Hope this helps!\nA. S=5\nB. X=-8\nC. Y=-3X+1\nD. It is a number that cannot be expressed as the ratio of two integers.\nE. A trinomial is a polynomial expression with three terms (after simplification).  The trinomial you are looking for is 3x^2+11x-4.\nF. 2 and (N^2 - 3). Factor out all common factors.\n\n\n\nA. Let K be the constant of variation.  T=KS. 12 =15K. K=4/5.  Now T=(4/5)S.  Substitue the values you've been given.  4=(4/5)S.  S=5\n\nB. X = -8.  An undefined expression includes division by zero, since this is impossible.  To make the denominator zero, X+8=0.\n\nC. Actually, it's Y=-3x+1.  This is just slope-intercept form, y=mx+b, where m=slope and b=y-intercept.  But think about it.  Y-intercept is where the graph crosses the y intercept, so the x-coordinate has to be zero. If you substitute 0 for x in the equation, you get Y=1, so this is correct.\n\nD. It is a number that cannot be expressed as the ratio of two integers.  It is still a real number, though (so no square roots of negative numbers!)  There are two types of irrational numbers: Radicals (the square root of five, the fifth root of 11) and Transcendental numbers (pi, e)\n\nE. A trinomial is a polynomial expression with three terms (after simplification).  For example, x^2 +x+1 or y^21+13y^3+231.  Each term has a different exponent of the variable (Including zero.  That is, y^0 =1)\nTo get the trinomial you are looking for, use the FOIL technique for multiplying binomials.  Multiply the First terms 3x*x (=3x^2). Outside terms are 3x*4 (=12x) and Inside terms are (-1)*x (=-x).  Finally, the Last terms together are (-1)*4 (=-4).  Now add all of these products together.  3x^2+12x+-x-4 = 3x^2+11x-4.\n\nF. Factor out all common factors.  Your original equation is 2n^2-6.  2n^2 = 2*n*n.\n6= 2*3.  \nThe only factor these have in common is two.  Take that out:\n2*(n^2-3)\n\n\n[edit] I just successfully finished an honors Algebra 2 course and I consulted my textbook to answer this question, so I know what I'm talking about!
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Please visit http://mathforum.org/library/drmath/view/52240.html
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http://mathforum.org/library/drmath/sets/high_calculus.html
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First you have to arrange it so the powers of x count down, which they already do.  I think you have to also get the leading coefficient = 1, but I'm not sure - better check on that.  It is a second-degree trinomial b/c there are 3 terms (tri=3, trinomial) and the highest power of x is 2 (2nd degree).
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milan.milanovic.org/math/english/pi/machin.html \n\nCheck this site
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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.
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well it depends for what exam. if you are looking for olympiad maths or higher level maths, i recommend www.kalva.demon.co.uk else you can try drmath.com if you have any queries on a question they do provide answers though sometimes their answers are not correct. so beware and have fun..
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Have a look here: http://www.coursework4you.co.uk/sprtlitra7.htm
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this page looks helpful:\n\nhttp://id.mind.net/~zona/mmts/functionInstitute/linearFunctions/lsif.html
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Probability function of a poission distribution\n==> P(X=x) = e^(-m)*m^(x)/x! \n[x= 0,1,2,.... ]  countably infinite\n\nMean = 4 ==> m = 4\n\nProbability that x = 2\n==> P(X=2)= e^(-4)*4^(2)/2!\n          = \n\nProbability that x <= 2\n==> P(X<=2)= P(X=0) + P(X=1) + P(X=2)\n==> P(X<=2)\n= e^(-4)*4^(0)/0! + e^(-4)*4^(1)/1! + e^(-4)*4^(2)/2!\n= e^(-4) {1/1 + 4/1 + 16/2}\n= e^(-4) {1 + 4 + 8}\n= e^(-4) {13}\n\nProbability that x > 2\n==> P(X > 2)\n= P(X=3) + P(X=4) + ...... \n(Or) = 1 - {P(X<=2)}\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/
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http://www.purplemath.com/modules/fcns2.htm
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Megami of Nekro is mostly correct, but I will try to clarify the situation a bit. \n\n1. factorize, or factor\n\nIf you have an algebraic expression that can be described as the product of some simpler expressions, to rewrite the expression as this product is to factorize it. \n\nSuppose you want to factorize the polynomial x^2 -3*x. In the case of polynomials of some degree, we regard polynomials of lower degree as simpler. In this case, the polynomial x^2-3*x is of degree 2, because the biggest power of x is 2 (the x^2 term). The lower degree polynomials in this case are of degree 0, called numbers, and of degree 1, called monomials. We can write the polynomial x^2-3*x as the product of lower order polynomials in the following way:\n\nx^2-3*x=x(x-3)\n\nThe two monomials x and (x-3) are factors of the polynomial x^2-3*x, and by expressing the polynomial as a product of two lower order polynomials, we have "factorized it" or "factored it". \n\n\nHere is another one:\n\nx^4+2x^2+1 is factorized as \n(x^2+1)(x^2+1)\n\nWith grade 8 mathematics, you cannot go any farther than this. We replaced a 4th degree polynomial with the product of its two factors, two 2nd degree polynomials (remember, we have decided that lower degree polynomials are simpler)\n\n\nAside: It is also interesting to note that one can factor whole numbers sometimes into a product of smaller whole numbers, which we regard as simpler. Therefore, \n\n10=5*2\n\nand 24=2*2*2*3\n\nThis is also called factoring. Numbers that can be expressed this way are called composite numbers. Numbers that cannot be expressed this way are called prime numbers (the real definition is slightly more complicated: you can look it up).\n\n2. Expand\n\nTo expand an expression is usually the opposite of factorizing. You multiply all the parts of the expression, at least as much as you can, using the distributive law, etc. \n\nIn the case of the expression x(x-3), we multiply our two monomials and return to the polynomial x^2-3*x:\n\nx(x-3)=x^2-3*x\n\nHere is another example. To expand\n\nx*(y-2)*(z+5)*y\n\nwe just multiply these factors together, \n\n x*(y-2)*(z+5)*y=\n(xy-2x)*(zy+5y)=\nxzy^2+5xy^2-2xzy-10xy\n\n\n\n3. Simplify\n\nThere is some ambiguity about what it means to simplify an expression. Computer algebra programs that simplify expressions sometimes use one definition of the simplest expression, and sometimes another. It all really depends on what shows up the features you are more interested in at the time. \n\nUsually to simplify an expression you have to do things like:\n\na. add up any like terms\n\nTo simplify  5x + 2y+ 2x+3+4, we add the terms with just numbers in them, and add the terms with just xs in them:\n\n5x + 2y+ 2x+3+4=7x+2y+7\n\nb. Multiply any numbers that can be multiplied\n\n5*2*x+2+y simplified is usually 10x+y+2\n\nc. Usually, when one simplifies an expression, the big powers are presented first, and the small powers follow (or sometimes vice versa).\n\nx+x^3+x^2 can be simplified as either \n\nx^3+x^2+x or x+x^2+x^3, depending on taste, the needs of the moment, etc.\n\n\nd. Usually when simplifying, any numbers or polynomials in ratios that can be divided will be\n\nSo 10/6 is often simplified as 5/3\n\nAnd (x^2-3*x)/x is often simplified as x-3\n\nHopefully that should get you started.
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try sosmath.com  they helped me before. if they don't have the answer you are looking for they have a list of links to tons of other math sites.  \n\nand there probably is someone on your campus that can help as well.  One on one help is always best...\n\nGood luck!