One factor of f(x)= x^3+x^2-22x-40 is x+4. Find the other factors..?
547775
f(x)= x^3+x^2-22x-40\nf(x)= (x+4)(ax^2 + bx + c)\n\nComparing coefficients,\nx^3: 1=a\nconstant: -40=4c => c=-10\nx^2: 1=b+4a => 1=b+4 => b=-3\n\nf(x) = (x+4)(x^2-3x-10) = (x+4)(x+2)(x-5)\n\nThe other factors are (x+2) and (x-5).
700407
f(x)=x^5-13x^3+36x \nf(x)=x(x^4 - 13x^2 + 36)\nf(x)=x(x^4 - 9x^2 - 4x^2 + 36)\nf(x)=x(x^2(x^2 - 9) - 4( x^2 - 9))\nf(x)=x(x^2 - 9) (x^2 - 4)\nf(x)=x(x-3)(x+3)(x-2)(x+2)\n\nas u havent mentioned domain or range of the function,,\nit is assumed ,,that domain is any real number,,\nthus,,the roots of ' x ' can be,,3,,-3,,2,,-2,,or any real number,,\n\nfoe example--\nf(1)=1(1-3)(1+3)(1-2)(1+2)\nf(1)=(-2) x (-4) x (-1) x (3)\nsoo,,\nf(1)=-24\n\nand ya,,f(x),,need not be zero always,,\nthis is what ppl r used to,,and soo,,they forgt that it is not zero always,,\nsoo,,\nzeros of the function here,,r,, any real number,,
650997
f(3(x - 4)) \n= f(3x - 12) \n= (3x - 12 + 1)^2 - 4 (3x  - 12)\n= (3x - 11)^2 - 12x - 48
977376
at the first 4 you use a^3-b^3=(a-b)(a^2+ab+b^2)\n and a^3+b^3=(a+b)(a^2-ab+b^2)\n\nfor the next 2 :\nthe first 2 factors and the second 2 factors have something in comon\n\nthe next ones use the above formulas plus one more \na^2-b^2=(a-b)(a+b)
886499
1). f(x) = (x-3)(x²+3x-5) +2. Multiply out to get your answer.\n2). f(1) =1, so we have to solve\nx³-8x²+17x-10 = 0.\nWe know x = 1 is a root of this equation, so x-1 is a\nfactor of the left-hand side. The other factor, obtained\nby long division or synthetic division is\nx²-7x+10.\nNow set this equal to 0 and solve:\n(x-5)(x-2) = 0,\nso the other 2 solutions are x = 5 and x = 2.
36762
Factor by grouping:\nx^2 (3x - 5) - 3 (3x - 5)\nAnswer: (3x - 5)(x^2 - 3)
900659
1. f(x) = 200x - 15/40 - 2x\n\nI assume you mean\nf(x) = (200x - 15)/(40 - 2x)\n\n= 5(40x - 3)/[2(20 - x)]\n= 5/2 * (40x - 3)/(20 - x)\n= 5/2 * (40x - 800 + 797)/(20 - x)\n= 5/2 * -40(20 - x)/(20 - x) + \n= 100 + 5/2 * 797/(20 - x)\n\nNow we can deal with it, as messy as it looks as before it was 'improper' since the degree of the numerator was the same as the degree of the denomonator and a rational function is 'improper' if degree numerator ≥ degree denominator\n\nNow lets look at 5/2 * 797/(20 - x)\n\nA. When x < 20, 20 - x > 0 so 5/2 * 797/(20 - x) > 0. \nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) > 100\nAND when x > 20, 20 - x < 0 so 5/2 * 797/(20 - x) < 0. \nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) < 100\n\nie for x > 20, f(x) < 100 and for x < 20, f(x) > 100\n\nB. When x = 20 20 - x = 0 so 5/2 * 797/(20 - x) is undefined.\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) is undefined.\nAs x → 20 from the left ,5/2 * 797/(20 - x) → +∞\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x))→ +∞\nAND as x → 20 from the right, 5/2 * 797/(20 - x) → -∞\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x))→ -∞\n\nie vertical asympotes at x = 20\n\nC As x → ±∞, 5/2 * 797/(20 - x) → 0\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) → 100\n\nie horitontal asymptotes at f(x) = 100\n\nD. You now have enough information to sketch this\n\n2. Work the function g(x)=x^2/x-2\n\nI assume you mean g(x)=x^2/(x - 2)\n\nAgain this is improper as deg numerator = 2 and deg denominator = 1 (< deg numerator)\n\nSo g(x) = x^2/(x - 2)\n= (x^2 - 4 + 4)/(x - 2)\n= ((x - 2)(x + 2) + 4)/(x - 2)\n= x + 2 + 4/(x - 2)\n\nDo steps A. to D as above\n\nA. Check out the signs of the denominator of the fraction\nWhen x < 2,  x - 2 < 0 so 4/(x - 2) < 0 \nTherefore g)x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)} < x + 2\n\nWhen x > 2,  x - 2 > 0 so 4/(x - 2) > 0 \nTherefore g(x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)} > x + 2\n\nie for x < 2, g(x) < x + 2 and for x > 2, f(x) > x + 2\n\nB. When x = 2,  x - 2 = 0 so 4/(x - 2) is undefined.\nTherefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) is undefined.\nAs x → 2 from the left, 4/(x - 2) → -∞\nTherefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) → -∞\nAND as x → 2 from the right, 4/(x - 2) → +∞\nTherefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) → +∞\n\nie vertical asymptotes at x = 2\n\nC. As x → ±∞, 4/(x - 2) → 0\nTherefore g(x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)}  → x + 2\n\nTherefore g(x) is assymptotic to g(x) = x + 2\n\nD. You now have enough information to sketch this
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.
484892
(x+3)(2x+4) is incorrect.\n(x+3)(2x+4) = 2x^2 + 10x + 12\nSolution:\nMultiply the coefficient of the x^2 term with the constant.\n2(12) = 24\nYou are looking for factors of 24 that add the 11 (which is the coefficient of the x term).\n(3)(8) = 24\n3 + 8 = 11\nReplace 11x with 3x + 8x\n2x^2 + 11x + 12\n= 2x^2 + 3x + 8x + 12\nFactor by grouping\n(2x^2 + 3x) + (8x+12)\n= x(2x+3) + 4(2x+3)\n= (2x+3)(x+4)\nAnswer: (2x+3)(x+4)\nYou can easily check the answer by Foil.\n(2x+3)(x+4) = (2x)(x) + (2x)(4) + (3)(x) + (3)(4)\n= 2x^2 + 8x + 3x + 12\n= 2x^2 + 11x + 12
1029303
if you are asking for the factors,it is\n(x-15y)*(x+15y)
517819
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
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).
809123
(x-2)(x+1)(x-1)-6(x+1)\nNow common factor is x+1\n(x+1)[(x-2)(x-1)-6)]\n(x+1)(x^2-3x+2-6)\n(x+1)(x^2-3x-4)\n(x+1)(x-4)(x+1)\n(x-4)(x+1)^2
323323
This is the best I could come up with:\n\n(a)  (x^2-4)(x^4+4x^2+16)\n      (x-2)(x+2)(x^4+4x^2+16)\n\n(b)  (x^3-8)(x^3+8)\n     (x-2)(x^2+2x+4)(x+2)(x^2-2x+4)\n\n(c)  If you take out the factors that are the same, what's left should be equal hence:\n     x^4+4x^2+16=\n(x^2+2x+4)(x^2-2x+4)\n\n(d)  x^(a^2)+2ax^a+4a^2=\n(x^a+ax+2a)(x^a-ax+2a)
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
760867
f(x) = (x+3)(x^2 + 16)
447887
You probably mean\np(x) = 2x^4 +5x^3 – 3x^2 -2x -6\nThe answer is e) Both b,d\nbecause when you multiply the factors you should get the original expression and (x+3)(2x^3 –x^2 -2)= 2x^4 +5x^3 – 3x^2 -2x -6\nThis was quickly seen because you need x^4 which is only obtainable by x multiplied by x^3 in your problem :)\nHope this helps