how do you factor x-x^5?
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.
35579
5(x^4 -9) = 5(x^2 + 3) (x^2 - 3)
779760
if by this you mean\n\nf(x) = (x^(2/5)) * 5(x - 1), then\n\nf'(x) = (2/5)x^((2/5) - 1)*(5(x - 1)) + (x^(2/5))(5(x - 1)')\nf'(x) = (10(x - 1)/5)x^((2/5) - (5/5)) + 5x^(2/5)\nf'(x) = (2(x - 1))x^(-3/5) + 5x^(2/5)\nf'(x) = ((2x - 2)/(x^(3/5)) + 5x^(2/5)\nf'(x) = ((2x - 2) + 5(x^(2/5) * x^(3/5)))/(x^(3/5))\nf'(x) = (2x - 2 + 5x^((2/5) + (3/5)))/(x^(3/5))\nf'(x) = (2x - 2 + 5x^(5/5))/(x^(3/5))\nf'(x) = (2x - 2 + 5x)/(x^(3/5))\nf'(x) = (7x - 2)/(x^(3/5))\n\n------------------------------------\n\nBut if you mean\n\nf(x) = (x^2)/(5(x - 1))\n\nf'(x) = (x^2)(5(x - 1))^(-1)\nf'(x) = (2x^(2 - 1))((1/5)(x - 1)^(-1)) + (x^2)((-1/5)(x - 1)^(-1 - 1))\nf'(x) = ((2x)/(5(x - 1)) + ((x^2)/(5(x - 1)^2))\n\nANS : f'(x) = ((2x)/(5(x - 1)) - ((x^2)/(5(x - 1)^2)\n\nor you can do it like this\n\nf'(x) = (2x(x - 1) - x^2)/(5(x - 1)^2)\nf'(x) = (2x^2 - 2x - x^2)/(5(x - 1)^2)\nf'(x) = (x^2 - 2x)/(5(x - 1)^2)\n\nor\n\nf'(x) = (x^2 - 2x)/(5(x - 1)(x - 1))\nf'(x) = (x^2 - 2x)/(5(x^2 - 2x + 1))\nf'(x) = (x^2 - 2x)/(5x^2 - 10x + 5)\n\nANS : (x^2 - 2x)/(5x^2 - 10x + 5)
1002101
factor 9x^2y^2-9y^2\n9y^2(x^2-1)\n9y^2(x+1)(x-1)
864177
1. Move the constant to the right side:\nx^2 - 6x = 4\n\n2. Take half of the x coefficient (in this case, -6), square it, and add it to both sides:\nx^2 -6x + 9 = 13\n\n3. Factor the left side to (x + a)^2, where a is the number you just squared:\n(x - 3)^2 = 13\n\n4. Take the square root of both sides, remembering that this will give you both a positive and a negative root:\nx - 3 = +/- sqrt 13\n\n5. Move the constant on the left to the right:\nx = 3 +/- sqrt 13
967957
There really isn't any way to do this other than guess and check. If you find that you substitute some value in and find f(a) = 0, then (x-a) will be a factor. You can repeat that a few times until you have found some roots, and then do a long division to something which you *can* solve by hand (a quadratic or linear).\nNote also that you only need to check things which are factors of the constant term, and there aren't too many.\nIn the first case, you will find that substituting in -2, -1 and 3 will all give 0, but you may have trouble finding the last root. So you will have to do some long divisions (divide by (x+2)(x+1)(x-3)), and you will end up finding that the -1 was a repeated root - so the answer is (x+2)(x+1)^2(x-3).\n\nSame thing for the second: you will find -1 as one root. You can then do a long division and find -1 is repeated, and again, and again, and again. The actual factorisation is (x+1)^5 - you may have noticed this if you knew much about pascals triangle and recognised the coefficients are an entire row - 1,5,10,10,5,1.
198477
x^2-6x+5=\n\nx^2-5x-x+5=  [presenting -6x as -5x-x]\n\nx(x-5)-1(x-5)= [getting out common factor]\n\n(x-1)(x-5)\n\nTo see that it was done correct put any small number in the original expression and the result and compare them. If you did right, you should get two numbers which are equal to each other.\n\nPutting x=1 on the left side(the original expression):\n\n1^2-6*1+5=1-6+5=-5+5= 0.\n\nPutting x=1 on the right side(the result):\n\n(1-1)(1-5)=0*(-4) = 0.\n\nWe got on both sides the number 0, as expected, which probably means we did it right.\nTher final answer is:\n\n(x-1)(x-5)\n\nHope that helps.
835153
.more kids homework... factoring?\n1. rewrite the middle term as the sum of two terms and then factor completely 15x^2 -21x +6\n\n3(5x^2-7x+2)\n3(5x^2-5x-2x+2)\n3(5x-2)(x-1)\n\n2. rewrite the middle term as the sum of two terms and then facto by grouping. x^2 - 14x +45 \n\nx^2-9x-5x+45\n(x-9)(x-5)\n\n3. rerewrite the middle term as the sum of two terms and then facto by grouping x^2 + 19x +90\n\nx^2-10x+9x+90\n(x+10)(x+9)\n\n4. rewrite the middle term as the sum of two terms and then facto by grouping x^2 -2x -63\n\nx^2-9x+7x-63\n(x-9)(x+7)\n\n5.solve x^2 = -6x\n\nx=-6
517819
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
590233
Hint:\nDo a simple substitution u = x-5\nWhich means x = u + 5\n\nYou should get\n\n(2/5)(x-5)^(5/2) + 2(x-5)^(3/2) + C
283601
1) First we move all the factors with x to one side, and the ones without x to the other.  Result will be:\n\n4x/3 - 2x + x = 3/2 + 1/5\n\n2)  Then, we factor out an x from the left side and get:\n\nx(4/3 - 2 + 1) = 3/2 + 1/5\n\n3)  Calculate the actual numbers on both sides\n\nx(4/3 - 6/3 + 3/3) = 15/10 + 2/10\nx(1/3) = 17/10\n\n4)  Divide both sides by 1/3 and get a final result of...\n\nx = (17/10)/(1/3)\nx = (10/17)*(3/1)\nx = 30/17\n\n5)  Comform that I PWN n00bs.  And... I'm pretty sure I pwn n00bs.  Therefor, we are done!  ^-^
402918
I think you wrote 10x*2yz - 22x*3yz.  To factor this oyu must take out like terms by dividing.  You can see that both terms share a 2, y, z, and an x*2.  After the  division you get  2x*2yz( 5 - 11x).  I hope this helps.
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
954896
x^2 + -2x = 5   Isolate the x^2 and x terms.\n\nx^2 + -2x = 5   Divide both sides of the equation by 1, which is the a-coefficient.\n\nx^2 + -2x + (-2/2)^2 = 5 + (-2/2)^2   Add half of the b-coefficient squared. In this equation that value is (-2/2)^2, and can be simplified to 1.\n\n(x + -2/2)^2 = 5 + (-2/2)^2   Write the x^2 and x terms side as a perfect square binomial.\n\n(x + -2/2)^2 = 5 + 1   Simplify the other side (simplify fractions).\n\n(x + -2/2)^2 = 6   Simplify the other side (combine numbers/fractions).\n\nSQRT[(x + -2/2)^2] = {+,-}SQRT[6]   Find the square root of both sides.\n\nWhen you combine and solve you'll get\n\nx = -1 plus or minus SQRT[6]
565133
We start by factoring the polynomial.  A little trial and error (testing the factors of 2, which is the constant term) reveals that 2 is a root.  So, we know that (x-2) is a factor.\n\nDividing the polynomial by (x-2) using synthetic division, we find the other factor is 6x^2+x-1.  This can be factored as (3x-1) (2x+1).\n\nSo, the equation, in factored form, is\n(x-2)(3x-1)(2x+1)=0\n\nSo, the roots are x=2, x=1/3, and x=-1/2.
219588
3x^3 + 2x^2 + x - 6\n= 3x^3 + 5x^2 - 3x^2 + 6x - 5x - 6\n= x(3x^2 + 5x + 6) - 1(3x^2 + 5x + 6)\n= (3x^2 + 5x + 6)(x - 1) \n\nThe factors are : (3x^2 + 5x + 6) and (x - 1) \n\nAlternatively, using polynomial division : \n(3x^3 + 2x^2 + x - 6) divided by (x-1) yeilds \n(3x^2 + 5x + 6)
689918
use  power rule and product rule\nf'(x)=[1/5]*5*x^4 + 2x(x^2-x-9)+ \n(x^2+9)(2x-1) + 0=\n\nx^4 + 2x^3 - 2x^2 -18x + 2x^3 - x^2 +18x -9 =\n\nx^4 + 4x^3 - 3x^2 -9