how do you factor polynomials that look like this: x^2 - 4?
914223
difference of squares rule:\n\nx^2 - y^2 \n\n(x-y)(x+y)
907035
You have \n100 +4X^2 -16y -40x\n\nFirst let's factor out a 4, since 4 is a factor of each term, that gives\n\n4*(x^2 -10X +25) -16Y\n\nOf course I rearranged the terms so the highest powers of X are\non the left.  I didn't factor 4 out of -16Y because I want the \nquadratic (X^2-10X+25), which I know I can factor.\n\nsince (X^2-10X+25) = (X-5)*(X-5) = (X-5)^2, we have\n\n4*(X-5)^2 -16Y\n\nThe secret is being able to recognize that \n(x^2 -10X +25) = (X-5)^2
1079811
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!
91753
factorize\nx^2 + x = x(x+1)\n\nfactorize\nx^2 - 1 = .... (you should know)\n\n2 = 2 (!!!)\nTake the lcm of the three denominators (each different factor (=parenthesys) once and with the highest power)\n\nIt is 2x(x+1)(...)   (fill the dots)\n\n.....\n\nnow if you can't go on, you need to open your book again and study, or pay more attention at school at your math classes, and ask immediately for further explanation to your teacher when you feel you've not understood.\n\nGood work and good study
594239
ok so you have p^2 + p, you can factor out p from both parts of the expression, it will end up like this: p(p + 1).  Hope that works.
764068
2x - 2 = x^2 - 4x + 3\nSet one side equal to 0\n0 = x^2 - 6x + 5\nFactor\n0 = (x - 5)(x - 1)\nSet each factor equal to 0\n0 = x - 5              0 = x - 1\nSolve\n5 = x                  1 = x
1108301
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)
408031
First, factor the polynomial using the Rational Roots Theorem. All possible factors of -36 are (+or-) 1, 2, 3, 4, 6, 9, 12, 18, 36.  Use synthetic division for each, until you find one root.  In my solution I got x - 4 as a root and the other x^3 + x^2 + 9x + 9.  Continue with the factoring.  Here I got x + 1 as a factor and the other x² + 9. Therefore, the factorization is:\nP(x) = (x - 4)(x + 1)(x² + 9)\n\nThe zeroes are all the roots when P(x) = 0\nTherefore,\n(x - 4)(x + 1)(x² + 9) = 0\nx = 4\nx = -1\nx² = -9 or \nx = 3i or -3i\n\n^_^
78119
I guess you must factorize them.\nAll the polynomials you wrote are to be factorized by partial factorization.\nI think you know the general rule.\nA(x)P(x)+A(x)Q(x) + B(x)P(x) + B(x)Q(x) = \n---> factorize between the 1°-2° and 3°-4° <---\nA(x)[P(x) + Q(x)] + B(x)[P(x) + Q(x)]\nYou get the same binomial term [P(x) + Q(x)] which is a common factor. Factorize:\n[P(x) + Q(x)]*[A(x)+B(x)]\n\nFor example, in the first:\nA()=a\nB()=b\nP(x)=2x\nQ(y)=3y\nYou get \na(2x+3y)+b(2x+3y) = (2x+3y)(a+b)\n\nIn all the others you must rearrange the terms to use the above steps:\nin xy+32+8x+4y the repeating patterns are 8x + xy and 32 + 4y ....\n\nThink about it and good work
464534
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.
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
1164880
If x-2 and x+1 is a factor, then P(2) = 0 and P(-1) = 0\nP(2) = 2^4 + a*2^3 - 2^2 +b*2 - 12\n= 16 + 8a - 4 + 2b - 12\n= 8a + 2b = 0\nP(-1) = 1 - a - 1 - b - 12\n= -a - b - 12 = 0\nYou now have a system of equations.\n8a + 2b = 0\n-a - b - 12 = 0\nSolve by substitution. a = 4 and b = -16\nNow use synthetic division or long division to get the other roots.\nx = 2 and x = -1 are given.  The other roots are -3 and -2.\n\n2) Not sure\n\n3) Use sythetic division or long division.\nYou will get that f(x) divided by (x+2) is 2x^2 - 2ax + 3a.\nIf -2 is the only real root, then the other roots must be imaginary.\nIf the roots of a quadratic equation are imaginary, then the discriminant is less than 0.\ndiscriminant = b^2 - 4ac, where a = 2, b = -2a and c = 3a\nb^2 - 4ac = (-2a)^2 - 4(2)(3a)\n= 4a^2 - 24a <0\nFactor:\n4a(a-6) < 0\nIf a<0 or a>6 then 4a(a-6) is positive.\nIf 0<a<6, then 4a(a-6) is negative.\nSo 0<a<6.\n\n4) Not sure, but I'll think about it.
670366
Yes.  There are algorithms for factoring polynomials up to 4th degree.  These algorithms are guaranteed to work, however in the case of the 5th degree polynomials and above, there is no algorithm that is guaranteed to find the solution.  In part this is because the roots can be irrational numbers.  I was proved by Abel in 1824 that no such algorithm can even exist.\n\nThere are iterative methods to factor polynomials of 5th degree and above, but they provide approximations, they are not guaranteed to find the correct answer.  The iterative methods are quite complicated and are equivalent to finding eigenvalues of matrices.\n\nIf you are in high school or early college, all the problems they will assign you should be solvable with either the linear, quadratic, or cubic equations.
439000
Since you cannot factor any binomial to form a monomial, the only factoring you can do is by "common monomial factoring". After that, cancel like factors, but if there are no common factors, that means you cannot factor it anymore.\nExample:\n\n(3x - 3)/(2x - 2)\nyou can get common monomial factors for each:\n3(x - 1)/2(x - 1)\ncancel the common factors\n3/2\n\n(4x² - 6x)/(2x + 10)\n= 2x(2x - 3)/2(x + 5)\n= x(2x - 3)/(x + 5)\n\n(x + 3)/(x + 5)\n---> this cannot be simplified anymore\n\n^_^
617264
Since each term has at least x in it, divide out the x by distributive property:\n\nx(x^2 - 4xy + 4y^2)\n\nThen factor the rest by FOIL:\n\nx(x-2y)(x-2y)
361536
Let's call f(x) = x^2 - 4\n\nThe graph of f(x) = x^2 - 4 looks like a smile ( U )  with the graph passing the y - axis at -4.\n\nThe roots are at x = 2 and x = -2.  This can be found by solving for x^2 - 4 = 0\nx^2 - 4 = 0\n(x + 2)(x - 2) = 0\nso x + 2 = 0, or x - 2 = 0\nso x = -2 or x = 2\n\nsince you're int'ing from x = -1 to x = 3, you have to break up the int into\n[x = -1 to x = 2, where f(x) is negative and take the absolute value] \nand\n[x = 2 to x = 3, where f(x) is positive]\n\nso int of f(x)\n= x^3/3 - 4x + c\n\nso int of f(x) from x=-1 to x=3\n= {+ve value of int of f(x) from x = -1 to 2}\n                +\n  int of f(x) from x = 2 to 3\n\nSeparately\nint of f(x), x = -1 to 2\n= | (2^3)/3  - 4(2) - [((-1)^3)/3  - 4(-1)] |\n= | (8/3) - 8 - [(-1/3) + 4] |\n= | (8/3) - 8 + (1/3) - 4 |\n= | (9/3) - 12 |\n= | 3 - 12 |\n= | -9 |\n= 9\n\nint f(x), x = 2 to 3\n= (3^3)/3  - 4(3) - [(2^3)/3  - 4(2)]\n= 27/3 - 12 - [(8/3) - 8]\n= 9 - 12 - (8/3) + 8\n= 5 - (8/3)\n= 7/3 (or 2 1/3)\n\nAdd them together and you get:\n9 + 2 1/3\n= 11 1/3 which is appromixately 11.333_
532215
Your question is to factor in the following form y=a(x-p)^2+q for the question y=x^2-8x+8. Okay, let's roll.\n\ny=x^2-8x+8\n = (x-4)^2 -16 +8   [using (a-b)^2 = a^2 -2ab +b^2]\n = (x-4)^2 -8\n\nTherefore, in the form  y=a(x-p)^2+q, a=1, p=4 and q= -8.