divide the first polynomial by the second polynomial   p(x)= x^4 + 3x^3 - x^2 + 11x - 4, f(x)=x+4 show work?
817835
x^3-x^2+3x-1\nx+4&#92;x^4+3x^3-x^2+11x-4\n      -(x^4+4x^3)\n              -x^3-x^2\n            -(-x^3-4x^2)\n                       3x^2+11x\n                     -(3x^2+12x)\n                                   -x-4\n                                 -(-x-4)\n                                         0\nx^4+3x^3-x^2+11x-4=(x+4)(x^3-x^2+3x-1)
76424
You can reduce you work by knowing that if your polynomial looks like  Ax^n + ... + B that if you have a factor that looks like (ax + b) that a has to be a divisor of A and b has to be divisor of B. That means that the only possibilitys in your example are\n(x + 1), (x - 1), (-x + 1) and (-x -1)\nSince (-x - 1) = -1(x+1) and (-x + 1) = -(x - 1)\nyou only have to look at two cases (x+1) and (X-1)\n\nTo find out if either of the two is a factor, you can do long division on the polynomial just like you would with regular numbers. If you don't know how to do that look at the link below for an example.\nI can't show you here because it won't format correctly.
650997
f(3(x - 4)) \n= f(3x - 12) \n= (3x - 12 + 1)^2 - 4 (3x  - 12)\n= (3x - 11)^2 - 12x - 48
715426
x^4 + 5x^3 - 30 / x^2 - 3\n\nThe important thing is that, especially with division, polynomials need place value so 0 coefficients apply.\n\nRewrite as follows\n\nx^4 + 5x^3  + 0x^2 + 0x - 30 / x^2 + 0x - 3\n\n.................... x² + 5x + 3\n................. ____________________\nx² + 0x - 3 )x^4 + 5x³ + 0x² + 0x - 30\n.................. x^4 + 0x³ - 3x²\n.................  ------------------ .....↓\n........................... 5x³ + 3x² + 0x\n........................... 5x³ + 0x² - 15x\n........................... ------------------ ....↓\n.................................... 3x² + 15x - 30\n..................... ............... 3x² + 0x - 9\n.................................... ------------------\n....................... ................... 15x - 21\n\nSo x^4 + 5x³  - 30 = (x²  - 3)(x² + 5x + 3) + 15x - 21\n(Or quotient is x² + 5x + 3 and the remainder is 15x - 21)
854837
Show it's a zero just means you sub x = -3 in the polynomial and get the result zero.  If that happens, you can divide the polynomial by \nx -(-3),  i.e. x + 3, and then find the zeros of the remaining expression.  Want to do it yourself?  Good.  Then don't scroll down to see what I've done.\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n-3      2     5     -6    -9\n\n              -6       3      9\n\n         2    -1     -3      0\n\nP(x) = (x + 3)(2x^2 - x - 3)\n       =  (x + 3)(x + 1)(2x - 3)\n\nZeros are -3, -1, 3/2
798648
(x+3)(x^2+2x+4)\n\nStart by taking the first term in the first parenthesis and multiplying it by each term in the second parenthesis:\n\nx*x^2 + x*2x + x*4 = x^3 + 2x^2 + 4x\n\nNow take the second term and multiply it by each term in the second parenthesis:\n\n3*x^2 + 3*2x + 3*4 = 3x^2 + 6x +12\n\nNow add all of that together\n\nx^3 + 2x^2 + 4x + 3x^2 + 6x +12\n\nand combine terms\n\nx^3 + 5x^2 + 10x +12
277221
Here is a proof by induction on p, which is probably what you are looking for.\n\nNote:\nTo label equations that I will refer to later, I will use a number in parentheses followed by a colon just before the equation, for example "(4): x = 0". To refer back to a labelled equation, I will use the equation number in parentheses, for example "by (4), blah blah".\n\n11^p - 4^p is divisible by 7 for any integer p>=1.\n\nProof:\n\nBase step:\n-----------------------\nIf p=1, then\n\n11^p - 4^p\n= 11 - 4\n= 7,\n\nso 7 divides 11^p - 4^p when p=1.\n\nInductive step:\n-----------------------\nFix some p>=1 and suppose the desired result holds for p, i.e. that 11^p - 4^p is divisible by 7. Using this, we want to show that 11^(p+1) - 4^(p+1) is divisible by 7.\n\nFirst, since we assumed 11^p - 4^p is divisible by 7, we know that\n\n(1): 11^p - 4^p = 7*k\n\nfor some integer k. Then\n\n11^(p+1) - 4^(p+1)\n= 11*11^p - 4*4^p\n= 7*11^p + 4*11^p - 4*4^p\n= 7*11^p + 4*(11^p - 4^p),\nnow substituting from (1),\n= 7*11^p + 4*7*k\n= 7*(11^p + 4*k).\n\nSo 11^(p+1) - 4^(p+1) is divisible by 7.\n\nTherefore by the principle of mathematical induction, 11^p - 4^p is divisible by 7 for every integer p>=1.\n\nEnd of proof.\n\nHope this helps :)
54176
The instruction is to write a polynomial function with the zeros 0, 1, and 1+i. Please keep in mind that when a zero consists radicals or imaginary numbers, the conjugate of that zero is also a zero. For exmaple, here 1+i is a zero. Then 1-i must also be a zero.\n\nThis problem is like working backwards. Instead of factoring the polynomial, you are asked to do it backwards, starting with the zeros. Since 0, 1, and 1+i and 1-i are zeros, you would expect the last step of the factoring to look something like this:\n\nx(x-1)[x-(1+i)][x-(1-i)]\n\nNow simplify this expression a bit. It becomes:\n\nx^2-x[(x-1-i)(x-1+i)]\n\nSimplify this expression even further and you get:\n\nx^4-3x^3+4x^2-2x
824178
You typed exponents just fine!\n\nx^3 - x^2 -x + 7 divided by x+1 is a "long division" type problem rather than factoring because x+1 doesn't go evenly into the polynomial\n\nThe answer is x^2 -2x + 1 + 6/x+1
888899
f(x) = x^2 + 3x\n\nf(2-a) = (2-a)^2 + 3(2-a)\n\nExpand (2-a)^2 to (2-a)(2-a) and distribute the 3(2-a)\n\nf(2-a) = (2-a)(2-a) + 6 - 3a\n\nf(2-a) = 4 -2a -2a + a^2 + 6 - 3a\n\nThe solution is:\n\nf(2-a) = 10 - 7a + a^2
920343
(4/3)[x^3/(2-x)]^(1/3)\n[(2-x)(3)x^2-x^3(-1)]/\n(2-x)^2=\n\n 4x[1/(2-x)]^(1/3)\n[3x^2(2-x)+x^3]/\n3(2-x)^2
491455
(x^2+4x+1)/(x-2) = x+2+3/(x-2)\n\nx  : x^2+4x+1-x^2-2x =2x+1\n2  : 2x+1 -2x-4 =3
760715
If it is divisible by x - 3, then (x - 3) is a factor of that polynomial, and if x - 3 = 0, x = 3.  \n\nIf we substitute x = 3 to the polynomial, then it must yield 0 (since 3 is a root).\nx³ + 3x² + kx - 24\n\nThus,\n(3)³ + 3(3)² + k(3) - 24 = 0\n\nThen,\n27 + 27 + 3k - 24 = 0\n\nTranspose,\n3k = -30\n\nThus,\nk = -10\n\nThe polynomial then becomes\nx³ + 3x² - 10x - 24 = 0\n\nWe then divide it by x - 3, to factor it out.  The easiest method goes like this\nx³ - 3x² + 6x² -18x + 8x - 24 = 0\n\nNow, we group\n(x³ - 3x²) + (6x² - 18x) + (8x - 24) = 0\n\nWe factor out from each group\nx²(x - 3) + 6x(x - 3) + 8(x - 3) = 0\n\nWe then factor out x - 3\n(x - 3)(x² + 6x + 8) = 0\n\nWe can factor the quadratic trinomial\n(x - 3)(x + 2)(x + 4) = 0\n\nTherefore, your polynomial is also divisible by (x + 2) and (x + 4).\n\n^_^
1079776
you can simply the first one to (x-a)(x-a) then you can divide that by (x-a) and it equals (x-a)\n\nDo you know synthetic division; cuz you can divide the second problem by (x+2)^2 if you mulitply (x+2)^2 out.
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
30225
These problems are talking about composite functions, which is where the results of one function are input into the second.  Usually it's written with a small circle in between (I'll use gof(x)) and read "g of f of x".  Remember that the function closest to the variable is the one that acts on it first.\n\nOne helpful tool to use when you can't figure out a problem is simplifying it to a problem you've done before.  For example, if f(x)= 3x and g(x)=x^2 ("x squared") then the composition gof(x) means take the results of f(x) and put them into g(x).  \nSo then gof(x) = (3x)^2 = 9x^2.\n\nYour problem is a bit more complicated, but we can use the same process:  \n\ngof(x) would be g(ax-1) so \ngof(x) = (ax-1)^2 + 3(ax-1) + 1\nUse the distributive property to expand this to\ngof(x) = a^2x^2 - 2ax + 1 + 3ax - 3 + 1\nand combine like terms to get\ngof(x) = a^2x^2 + ax - 1   *\n\nSo far, we've done everything just from the definitions of g(x) and f(x).  Now we use the last piece of information we were given: \nthat gof(x) = 4x^2 + ax + b  *\nWe compare the coefficients on like terms of these two polynomials (with the stars) to determine that since a^2 must equal 4, \na = 2 and b = -1 (from comparison of the constant terms).\n\n\nSee if you can use a similar process for the second problem!
1047514
x^3 + 5x^2 - 8 divided by x -1 is:\n\nx^2 + 6x + 8\n\nor...\n(x + 2) (x + 4)\n\ndon't listen to byh. sorry I can't show you HOW to do the problem. I just couldn't figure out how to in type.