Using the Rational zeros theorem, list all POSSIBLE zeros: P(x)=6x^4-x^2+2x+4?
822259
If p/q is a rational root (and p, q have no common factor) then p divides the free term (4 here) and q divides the leading coefficient (6 here). Thus you can all the possible rational roots.
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
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)
791694
The rational roots of any polynomial:\n\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) \n\nare in the form ±p/q, where p divides a(0) and q divides a(n).\nIn your case, a(0) = -1 and  a(n) = 1, so the only rational roots possible are ±1.\nSubstitute ±1 into the equation to see if either is a root. (neither is)\n\nExample:\nx^3 + 4x^2 + 2x - 4 = 0\nThe only possible rational roots are ±1, ±2, ±4 (These numbers divide a(0), and only 1 divides a(3)=1 ).  Substituting each of these into the equation (synthetic substitution works well), we find -2 the only rational root.\n\n*****************************************\nProof by contradiction.\nLet the rational number p/q be a reduced root (i.e, q  and p are relatively prime integers) to the rational polynomial equation:\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) = 0\nand such that either p does not divide a(0) or q does not divide a(n).  [Note: without loss of generality we can assume all coefficients are integers by multiplying both sides by the least common multiple of the denominators of all the a(n)'s]\n\nThen we know that \na(n)p^n/q^n + a(n-1)p^(n-1)/q^(n-1) + ... + a(1)p/q + a(0) = 0\nmultiply both sides by q^n to get:\n\na(n)p^n + a(n-1) qp(n-1) + a(n-2) q^2p^(n-2) + ... + a(1) q^(n-1)p \n+ a(0)q^n = 0\n\nor a(n)p^n + q * f(a,p,q) = 0 for some function f of a,p,q.\nand a(0)q^n + p * g(a,p,q) = 0 for some function g of a,p,q.\n\nThis means \na(n)p^n = -q * f(a,p,q)\na(0)q^n = -p * g(a,p,q)\n\nBy the first equation, q must divide a(n) * p^n.\nBut q has no factor in common with p, so it cannot have a factor in common with p^n, so all the factors of q (ie, q itself) must divide a(n), and we assumed that q does not divide a(n), so p must divide a(0) * q^n (the only possibility left)\nBut likewise with the second equation, p must divide a(0) * q^n. But p does not have any common factors with q, so it cannot have common factors with q^n, so p must divide a(0) and we assumed that p does not divide a(0), so a contradiction arises.\nTherefore no such p/q can exist, therefore the roots are irrational.\n\nQ:\nwhy shouldn't q divide a(n) or p divide a(0)?\n\nBecause otherwise if both did then the roots COULD be rational (p/q could be a root).  If thats the case you prove/disprove it by example.  In other words, its easy to show that in your example the roots are all irrational because the only POSSIBLE rational ones are 1 and -1.  So you just substitute them in to see that they're not roots. My proof just shows that IF there are rational roots, they must be p/q where p|a(0) and q|a(n).\n\nThere is no general way to determine  by inspection that a random polynomial has all irrational roots. The only way is to check the POSSIBLE rational ones by substitution.
527455
If those are the zeros, you just do it like this for the first one:\n\ny=(x-5)(x-3i)(x+3i)\n\n\nIf your teacher wants it in general form, you have to multiply out all the terms. \n\ny=(x-5)(x^2+9)\n\ny=x^3+4x^2-45\n\n\nThe second set doesn't make sense because you have an i without a partner. In polynomial equations, the complex factors always come as conjugate pairs, like the -3i and +3i.\n\nAnyway, I hope this helps. Basically, you set up a bunch of factors that are x-each root.\n\nI just changed it because I realized that the 1. meant problem number 1.
712942
x^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1 = 0\nx^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1 - 10x^3 = 0\nSo (x - 1)^6 - 10x^3 = 0\nSo ((x - 1)²)³ - (³√10x)³ = 0\n\nSo ((x - 1)² - ³√10x)((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0\n\nSo ((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0 N0 real roots (Min value near x = 0.2 just more than 0.87)\n\nOR (x - 1)² - ³√10x = 0 \nie x² - (2 + ³√10)x + 1 = 0\nie x = ½[(2 + ³√10) ± √(4 + 4³√10 + ³√100 - 4)\n= ½[(2 + ³√10) ± √(4³√10 + ³√100)]\n\nSOLUTION\nx = ½[(2 + ³√10) ± √(4³√10 + ³√100)]
408031
First, factor the polynomial using the Rational Roots Theorem. All possible factors of -36 are (+or-) 1, 2, 3, 4, 6, 9, 12, 18, 36.  Use synthetic division for each, until you find one root.  In my solution I got x - 4 as a root and the other x^3 + x^2 + 9x + 9.  Continue with the factoring.  Here I got x + 1 as a factor and the other x² + 9. Therefore, the factorization is:\nP(x) = (x - 4)(x + 1)(x² + 9)\n\nThe zeroes are all the roots when P(x) = 0\nTherefore,\n(x - 4)(x + 1)(x² + 9) = 0\nx = 4\nx = -1\nx² = -9 or \nx = 3i or -3i\n\n^_^
751294
3(2x+4)=6(x+2)\n6(x+2)=6(x+2)\ninfinite solutions
785246
x^2-16=0\n(x+4)(x-4)=0\nx+4=0\nx=-4\n\nx-4=0\nx=-4\n\nx^3+14x^2+40x=0\nx(x^2-14x+40)=0\nx(x-10)(x-4)=0\nx=0\n\nx-10=0\nx=10\n\nx-4=0\nx=4\n\nx^4-5x^2+4=0\n(x^2-4)(x^2-1)=0\n(x-2)(x+2)(x-1)(x+1)=0\nxi2=0\nx=2\nx+2=0\nx=-2\nx-1=0\nx=1\nx+1=0\nx=-1
708917
(x^4) + (4x^3) + (2x^2) - 8x - 8=0\n\nGeneral:  ax^4+bx^3+cx^2+dx+e=0\nif p/q is a rational root then p is divisor in a and q is divisor in e =>\n\nin this case:\np belong to {-1,+1} & q belongs to {-8,-4,-2,-1,1,2,4,8} =>\n\np/q belong to {-8,-4,-2,-1,1,2,4,8}\n\nWe try  -2 as a root and see\n2^4 + 4*2^3 + 2 * 2^2 - 8 * 2 - 8=0\n\nWe can now do polynomial division by (x+2):\n\n(x^4) + (4x^3) + (2x^2) - 8x - 8 : (x+2) = x^3 +2x^2 -2x - 4\nx^4 . . . + 2x^3\n- - - - - - - - - -\n0 . . . . . . 2x^3\n. . . . . . . .2x^3 . .+4x^2\n- - - - - - - - - - - - - - - \n. . . . . . . . .0. . . -2x^2 . . -8x\n. . . . . . . . .  . . . -2x^2 . . -4x\n- - - - - - - - - - - - - - - - - - - -\n. . . . . . . . . . . . . . . . . . .-4x. -8\n. . . .  . . . . . . . . . . . . . . -4x  -8\n- - - - - - - - - - - - - - - - - - - - - - -\n. . . . . . . . . . . . . . . . . . . . . .  0\n\nexactly the same procedure as before gives x= -2 is root so now we go again\n\nx^3 +2x^2 -2x - 4 : (x+2) = x^2-2\nx3^. .2x^2\n- - - - - - -  - -\n. . . . . . .0 . .-2x -4\n. . . . . . .0 .  -2x -4\n- - - - - - - - - - - - -\n. . . . . . . . . . . . .0\n\nthe last term x^2-2=0 => x=±√ 2\n\nnow we have :\n\n(x^4) + (4x^3) + (2x^2) - 8x - 8 =\n(x+2)^2 * (x-√ 2) * (x+√ 2) = 0\n=======================\nso x = { x in R I x = -2 or x= -√ 2 or x= +√ 2}\nand x= -2 is root 2 time\n=================================
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
570705
Factoring out X gives:\nX(-X^2 + X + 6)=0\nUsing the quadratic formula to factor the terms inside the brackets gives:\n-X^2 + X + 6 = (X+2)(-X+3)\nSo we have:\nX(X+2)(-X+3)=0 and the zeros are X=0, X=-2, X=3
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
820806
-2x^2+6x-3=0\n=->2(x^2-3x+3/2)=0\n=>2(x^2-2x.3/2+3^2/2^2)-3/4=0\n=2(x+3/2)^2-(sqrt3/2)^2=0\n=>-2(x+3/2+sqrt3/2)(x+3/2-sqrt3/2)=0\nTherefore,x=-3/2-sqrt3/2 or -3/2+sqrt3/2
730483
Note that the function is zero when x=1, this means that you can factor out an (x-1)\n\nx^4-x^2+4x-4 = (x-1)(x^3+x^2+4)\n\nNote that x^3+x^2+4 is zero when x=-2 so you can factor out an (x+2),\n\nx^3+x^2+4 = (x+2)(x^2-x+2)\n\nThe quadratic, x^2-x+2, does not factor with real numbers. (Try out the quadratic equation to confirm.) So the answer is\n\nx^4-x^2+4x-4 = (x-1)(x+2)(x^2-x+2)
889891
~..-8,-6,-4,-2,0,2,4,6,8