Can you factor 8x^4 + 52x^3 + 66x^2 + 31x + 5??
205504
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
33423
i got it!\n\nit's 3(2c+7)(c-4d)(c+d)!!!!\n\n6c^3 + 21c^2 - 18dc^2 - 63cd - 24cd^2 - 84d^2\nfirst we take out factor 3:\n3(2c^3 + 7c^2 - 6dc^2 - 21cd - 8cd^2 - 28d^2)\nthen we take out (2c+7) from each pain inside the parenthesses:\n3[(2c+7)c^2 - (2c+7)3cd - (2c+7)4d^2)]\nwe take (2c+7) out:\n3(2c+7)[c^2 - 3cd - 4d^2]\nwe factor [c^2 - 3cd - 4d^2] trinomially:\nc^2 - 3cd - 4d^2 = (c-4d)(c+d)\nthen we get:\n3(2c+7)(c-4d)(c+d)\n\nam i right??\n\n:) yeah i didn't believe myself...\nbut if u really write down all your factors, it's much more easier to take them out. and it WAS fun :))
231046
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
234031
It's called factoring by grouping. Take a factor out of two of the four terms (and the same for the other two) so they share a common factor.\n(2x^3 - x^2) - (18x - 9)\nx^2(2x - 1) - 9(2x - 1)\n(x^2 - 9)*(2x - 1)\n(x + 3)*(x - 3)*(2x - 1)\n(ungkog WAS correct)
162631
yes u can\nt^2(s^4-4)is the factored form\nt^2 is the highest common factor\n\n\n75x^2 - 147y^2\n3 is the common factor\n3(25x^2-49y^2)\n3(5x^2-7y^2)(5x^2+7y^2) is the factored form of 75x^2 - 147y^2
397013
Look in your text book under "factoring polynomials"... or you could try this web site that explains this fairly thoroughly...\n\nhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Factoring.html
395786
Notice that both terms have a 'factor' of (p+q) \nThe first term is (p+q)*(p+q) and the second is (5)*(p+q)\nSo 'factor out' the (p+q) from each :\n= (p+q) * [ (p+q) + (5) ]\n= (p+q) * (p+q+5)\nis the factorization.
526134
You can do this using "synthetic division"\n....... ______________________\nx - 5 ) x^4 - 0x^3 - 3x^2 + 0x + 6\n\nI'm going to simplify this leaving out the x-terms and only showing the coefficients, but just imagine they are there:\n\n.............. 1 . 5 .22 110\n1 -5 ) 1 . 0 .-3 . 0 . 6\n......... 1. -5\n......... ------\n.............. 5 .-3\n.............. 5 -25\n.............. -------\n................... 22. 0\n....... ........... 22 -110\n....... ............----------\n....... ........ ........ 110  6\n....... ........ ........ 110 -550\n....... ........ ........ ------------\n....... ........ ........ ....... 556\n\nSo if I did it correctly the answer should be:\n1 5 22 110 remainder 556\n\nThis equates to:\n(x^3 + 5x^2 + 22x + 110) + 556/(x-5)
217152
Factor each one.\n(x^2-3x-18) / (x^2-4x-32) divided by (x^2-2x-24) / (x^2-x-20)\n= (x^2-3x-18) / (x^2-4x-32) times (x^2-x-20)/(x^2-2x-24)\n= [(x-6)(x+3)]/[(x-8)(x+4)] times [(x-5)(x+4)/(x-6)(x+4)]\n= [(x-6)(x+3)(x-5)(x+4)] / [(x-8)(x+4)(x-6)(x+4)]\nYou can now cancel anything on top with anything on bottom\nx-6 and x+4 cancels\n= [(x+3)(x-5)]/[(x-8)(x+4)]
658768
(x+5)(2x^2+4x+2)\nx(2x^2+4x+2)+5(2x^2+4x+2)\n2x^3+4x^2+2x+10x^2+20x+10\n2x^3+14x^2+22x+10
285479
You can factor out 15bc^2\n15bc^2(5b + 4c^4)
348539
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
243887
You are only allowed to factor if the equation is equal to zero; your's is equal to 4.  You need to solve "x^3 - 6x^2 +9x - 4 = 0"  using synthetic division.
669034
8x^2 -5 -3x^2 + 7\n8x^2 - 3x^2 -5 + 7\n5x^2 + 2
293617
(x+10)(x-10)\n(x-4)(x²+4x+16)\n(x+w)(x²-wx+w²)\n\nUnfortunately you just have to memorize the formulas.  There is no other way to figure out the factoring of the cubic expressions.
573287
factors\n1=1\n2=2x1\n3=3x1\n4=2x2=4x1\n5=5x1\n6=3x2=6x1\n7=7x1\n8=2x2x2=4x2\n9=3x3=9x1\n10=5x2=10x1\nMultiples\n\n1=1,2,3,4,5,6.......\n2=2,4,6,8,10,\n3=3,6,9,12,15\n4=4,8,12,16\n5=5,10,15,20\n6=6,12,18,24\n7=7,14,21,28\n8=8,16,24,32\n9=9,18,27,36\n10=10,20,30,40
604009
not to worry noname ,i'll do it for u:\nsee 25x^2 + 55x+10 times 2x^2 means u have to multiply them both ok?\nso it becomes 50 x ^ 4 + 110 x ^3 + 20 x ^2 \n= 10 x ^2 ( 5 x^2 + 11 x + 2)\n= 10 x ^2 (5 x^2 + 10x + x + 2)\n= 10x^2 (5x(x + 2) + 1 (x +2)\n= (10x^2) (5x + 1 ) (x+2)\nso this will be the simplied form.if u didn't understand it yet and need any help then u can contact me . alright!