Hi does anyone know where I can find some fee help with factoring polynomials?Your help is appreciated:)?
917477
where else?\n\nhttp://www.factoring-polynomials.com/
1022192
i have  a book about algebra..... and some website...see if they could help u
894899
Horners' method of synthetic division is used to find the roots of polynomials. \nwe have to know one of the roots to proceed with this. we can check if 1or -1 are roots of the polynomial using some easy tricks. \n1) if the sum of all the coefficients is equal to 0, then 1 is a root\n2) if the sum of coefficients of even terms is equal to the sum of the coefficients of the odd terms, then -1 is a root.\nnow, for the first equation, we already know that -2 is a root, because the question is to find the quotient when the polynomial is divided by x+2. \nin synthetic division, u first put down a 0 beneath the first coefficient, ie., 3. now add these two. the result is 3. \nnow multiply 3 with the root, ie., -2. the result is -6. write this beneath the second coefficient, ie., 2. the result is -4 after addition. \nrepeat the same steps until u reach the last coefficient.\nnow next we write 8 below -1. result is 7.\nnext we write 7*-2 ie., -14 below 5. the result is -9.\nlastly, we write -9*-2=18 below . the result is 12, but the result should be zero.\nso, the given equation is not divisible by x+2.\n\ncoming to the second equation, follow the same procedure as above and u will get the resultant coefficients at each stage to be 2,6,22,61 respectively. so it is not divisible by 3
383816
!!, !!!!!!!! are "multifactorials". They are described in the first attached article.\n\nBriefly,\nn!! = 1 for n=0 or 1,\nn!! = n * (n-2)!! for n>=2.\n\nn!!...! (k "!" signes) = 1, for n<k\nn!!...! (k "!" signes) = n * (n-k)!!...! (k "!" signes)\n\n(2n)!! = n! * 2^n\n(kn)!!...! (k "!" signes) = n! * k^n\n\nSo, we have\n1) 6! = 1*2*3*4*5*6 = 720\n2) 6!! = 2*4*6 = 48\n3) 0! = 1\n4) 5!! = 1*3*5 = 15\n(5!!)!! = 15!! = 1*3*5*7*9*11*13*15 = 2027025\n\nOther factorials are too huge to compute precisely, but we can evaluate them using Stirling formula (second article)\nln n! ~= n*(ln n) - n + (ln(2*pi*n))/2 + 1/(12*n)\n\n5) 5! = 120\n(5!)! = 120! =~ e^(120*ln(120) - 120 + (ln(2*pi*120))/2 + 1/(12*120)) =~ e^457,81 =~ 6.69*10^198\n6) 1234!!!!!!!! = 1234 * 1233 * 1232!!!!!!!! =\n= 1521522 * (154*8)!!!!!!!! =\n= 1521522 * 154! * 8^154 = e^972.43 =~ 2.1*10^422\n\nThus, 6!! - 6! + (5!)! - 1234!!!!!!!! - 0! + (5!!)!! =\n=~ 48 - 720 + 6.69*10^198 - 2.1*10^422 - 1 + 2027025 =~ -2.1*10^422\n\nAnswer: about -2.1*10^422
455652
By the remainder theorem, substituting a value b into a polynomial gives the remainder if you divided the polynomial by (x - b). So if you can find a value that goes into the polynomial that gives a value of 0, then you know that (x - b) is a factor, since there is no remainder. \n\nNow (x - b) can only be a factor of the polynomial if b is a factor of the constant at the end, in this case 5. So the possibilities to test are 5, -5, 1 and -1.\n\nTesting +/- 1 first as they are easiest\n\nPutting x = 1 gives 8 + 52 + 66 + 31 + 5 = 162.\nPutting x = -1 gives 8 - 52 + 66 - 31 + 5 = =4\nSo (x - 1) and (x + 1) aren't factors.\nPutting x = 5 clearly gives a positive value\nPutting x = -5 gives 0, so (x + 5) is factor.\n\nThe next step is long polynomial division.\n\n(x + 5)   )    8x^4 + 52x^3 + 66x^2 + 31x + 5\n\n8x^4 divided by x gives 8x^3 \n8x^3 * (x + 5) = 8x^4 + 40x^3\nSubtract this from the polynomial to find the remainder:\n12x^3 + 66x^2 + 31x + 5\n\nThe quotient so far is 8x^3\n\nRepeat this process to find the final quotient of:\n\n8x^3 + 12x^2 + 6x + 1\n\nIf you can recognise that this is of the form of a cubic:\n\na^3 + 3.a^2.b + 3.a.b^2 + b^3\nwhere a = 2x and b = 1, you have your solution.\n\nIf not, we again have to try to factorise by testing factors. Now it should be obvious that (x - b) can no longer be factor, since we know (x + 1) and (x - 1) can't be factors. We need to try something of the form (ax - b), by observing that x^3 has a coefficient of 8, so (2x + 1) is a reasonable factor to try, since 2 is a factor of 8 at the start of the polynomial and 1 is a factor of 1 at the end of the polynomial. You can test if it's factor, because if (2x + 1) is factor, then x = -1/2 must be solution to the polynomial being equal to 0, which it is.\n\nPerforming polynomial division again gives 4x^2 + 4x + 1, with the factors now at (x + 5)(2x + 1). From here you can use trinomial factorising techniques to get the final solution of (x + 5)(2x + 1)^3
913123
I think you can try this website \nhttp://www.algebra-test.com \nhttp://www.algebraavenue.com\nhttp://www.algebra-online.com/algebra-equation.html \nhttp://www.algebrahelp.com\nhttp://www.algebra-net.com\nhttp://www.algebra-online.com/solve-algebra-equation.html\nhttp://www.algebrahelp.com/lessons/equationbasics/pgw.htm\nhttp://www.algebrasolver.com\nhttp://www.edhelper.com/algebra.htm\nhttp://www.math.com/students/worksheet/algebra_sp.htm
178199
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.
362703
Free Video lessons!   \nForum/Help Center  Ask for help and help others \nand much more\nhttp://www.algebra.com/
154296
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.
375277
Mathworld.com.  Or, if you type the subject your interested in followed by "Forum" in google, you'll find some good help that way.  Fore xample, "calculus forum".  Also, you can ask me whatever you want.
26326
A procedure for finding the quadratic factors for the complex conjugate roots  of a polynomial P(x) with real coefficients.\n\n[x-(a+ib)][x-(a-ib)]=\nx^2+2ax+(a^2+b^2)=\nx^2+Bx+C.
500175
Quadratic polynomials are usually in the form ax^2 + bx + c. They are usually factorable (which means you can find it's simplified roots). These roots should look like (x-F) and (x-G) where F and G are numbers so that when multiplied together using FOIL (First Outside Inside Last), you get your original polynomial. The roots are also where the polynomial = 0, assuming you've set it to zero. If the polynomials do not factor nicely, then you have to use this formula:\n\n-b +/- root(b^2-4ac)\n--------------------\n      2a\n\n\nWhich will give you the roots. Sometimes you get imaginary roots where in the equation above you will be taking the root of a negative number.\n\nHope that helps.
964615
Please type smith's algorithm for polynomial matrices at yahoo search you get links hope fully one of them should provide you with a solution.
562954
You simply divide each term in the polynomial by that integer.. For your example here it goes..\n\n(3x^2+8x-2) divided by 2 \n\n> {3(x^2)/2} + {8x/2} - {2/2}\n\n>> Answer: 1.5 x^2 + 4x -1    or   (3/2)x^2 + 4x -1
445887
you have to take out an x because there are x's that are there that are common to both terms. x(1-x^4) then -x(x^4-1)\nso factoring is easier. Then -x(x^2+1)(x^2-1). Then it becomes -x(x^2+1)(x+1)(x-1). That's the answer.
895063
Google the word algebra help.  Purple math is great, but there are other sites also.  I am sorry I do not have the names at the moment.
32760
You can use the AC Method to answer this question:\n\nFirst, rewrite the expression so it's in the form Ax^2 + Bx + C where A,B, and C are constants (numbers)\n\n16x^2 + 8x - 15\n\nThere's no common factors we could take out first, so...\n\nMultiply A x C, which in this case is 16 x -15 or -240.  Then look for factors of this number that add up to B.  You're looking for two numbers that fill in the blanks below:\n__ x __ = -240\n__ + __ = 8\n\nPrime facorization will help give you some ideas, or make an ordered list of numbers that multiply to -240.  \n240 and -1\n120 and -2\n80 and -3\n60 and -4\n48 and -5\n40 and -6\n30 and -8\n24 and -10\n20 and -12\n\nSee how the last pair of numbers satisfies the conditions?\n\nRewrite your polynomial, substituting the 2 values in for B to have 4 terms:\n\n16x^2 + 20x - 12x -15\n\nThen factor by grouping\n(16x^2 - 12x) + (20x - 15)\n4x (4x - 3) + 5 (4x - 3)\n(4x + 5)(4x - 3)\n\nThis method works for most polynomials of the form Ax^2 + By + C!  \n\nYou can check your answer of (4x + 5)(4x - 3) by using the FOIL method and see that it does work.\n\nGood luck!