Factoring 8x^4 + 52x^3 + 66x^2 + 31x + 5?
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
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
200291
The factor common to the first 2 terms in #1 is 3x^4 because both terms are divisible by that (take the GCF of the numbers in front, and the lower of the two exponents).\n\nFor the second pair, the common factor is 4\n\nSo you'd get 3x^4(3x + 2) - 4(3x - 2) if you have to actually factor them\n\nProblem 2, the common factor of the first 2 terms is 4x^3 and for the second pair, 1.\n\nSo you'd get 4x^3(5x + 4) - 1(5x + 4)  which becomes\n\n(4x^3 - 1)(5x + 4).
998496
x^3-3x^2+10x-30 R 10\nit's x+3 / x^4+0x^3+x^2+0x-80\n              x^4+3x^3\n             -------------\n                    -3x^3+x^2\n                    -3x^3-9x^2\n                    -------------\n                             10x^2+0x\n                             10x^2+30x\n                             --------------\n                                      -30x-80\n                                      -30x-90\n                                     ------------\n                                              10
767628
I'm assuming you want to factor this\n\n5z^2 + 4z -12 = (5z -6)(z+2)
210100
Factor them out. \n2 = 2\n3 = 3\n4 = 2*2\n5 = 5\n6 = 3*2\nLowest common multiple = 2*2*3*5 = 60\nYou have 1 left over, so 60+1\n61.\n(2*30)+1, (3*20)+1, (4*15)+1, (5*12)+1, (6*10)+1
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
652953
I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
714160
x^2+6x+8/x^2-4-21 / x^2-4/x^2-7x\n=x^2 + 6x +(8/x^2) - 4 - (21/x^2) - (4/x^2) - 7x\n=x^2 + 6x -4 -7x +(8/x^2) - (21/x^2) - (4/x^2) \n=x^2 + 6x -4 -7x +[8-21-4]/x^2\n=x^2 + 6x -4 -7x -17/x^2\n=(x^2)*[x^2 - x -4  -17]/x^2\n=(x^2)*(x^2) - (x^2)*(x) -4*(x^2)  -17]/x^2\n=(x^4) - (x^3) - 4(x^2)x -17]/x^2
707418
20x^2+60x + 45\n= 5(4x^2+12x + 9) \n= 5((2x)^2+2(2x)3+3^2) = 5(2x+3)^2\n\n3x^2+12x+12 = 0\n=> x^2+4x+4 = 0 (divide by 4)\n=>(x+2) ^2 = 0 or x = -2, -2 (double root)\n\n7x^2-28 =0\n= x^2 = 4\n=> x = 2 or -2\n\narea = xy = 5*2 = 10cm^2
959197
I have factored it please refer to.\n\nFrom this solution can be found
798535
x^3 - x^2 + 5x + 5\n\n(x^3 - x^2) + (5x - 5)\n\nx^2(x - 1) + 5(x - 1)\n\nsince both x^2 and 5 have the same factor, you can factor it to\n\n(x^2 + 5)(x - 1)\n\nif you want, you can either leave it that way, or\n\n(x + sqrt(5i))(x - sqrt(5i))(x - 1)\n\nwhat you can probably just leave it as (x^2 + 5)(x - 1)\n\nyou can also try www.quickmath.com, click Factor under Algebra, then type in x^3 - x^2 + 5x - 5, then click enter. \n\nAnymore Questions Ask ME
930643
factor:\n\n(2x^2+x-6)/(2x+4)=\n\n(2x-3)(x+2) / 2(x+2)\n\nthe (x+2)s cancel out leaving\n\n(2x-3)/2
836052
You can start by factorizing 3x^4 from everything\nm(x) = 3x^4 (2x^3 + 5x^2 - 3x - 3)\n\nI think there's something wrong with your problem, none of the possible roots given by the Ruffini's rule are working. Perhaps, if instead of 6x^7 + 15x^6 - 9x^5 - 9x^4 it was 15x^7 + 15x^6 - 9x^5 - 9x^4, then it could be factorized very nicely in\nm(x) = 3x^4 (x+1)(5x^2-3) \nor if instead you wrote 6x^7 + 6x^6 - 9x^5 - 9x^4 then it'd factorized to\nm(x) = 3x^4 (x+1)(2x^2-3)
930308
x=[2+/-rt(4+60)]/2\n=[2+/-8]/2\n=-3 or 5\nso the factors are\n(x+3)(x-5)
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
263286
the problem you presented above can be represented as,\n\n(8x^8/2x^2) + (16x^6/2x^2) - (12x^3/2x^2)\n\nthen you can simply divide each of the term\nyou'll get\n\n4x^6 + 8x^4 - 6x