Using Rational Zeros theorem?
477858
There are 2 factors of the leading coefficient 2, and 3 factors of the constant 4, so 2 times 3 is 6.  Then there are 2 possible signs, so there are 12 possible zeros.  But that includes duplicates.  The possibilies are ±4/1, ±2/1, ±1/1, and ±4/2, ±2/2, ±1/2.  2/1 = 4/2 and 2/2 = 1/1, so if you remove the duplicates there are 8 possibilities.
680780
You can use infinite series...\n0 = 0+0+0+...\nsubstitute 0 = (1-1)\nso \n0 = (1 - 1) + (1 - 1) + (1 - 1) + ...\n\nuse associatve law of addition so that\n0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... \n\nthen cancel out leaving you with \n0=1\n\nerror: using associative law on on infinite sum that does not converge
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^_^
561184
a)\ni: s may not be zero, as m/s would be undefined\nii: s=m/(x-z)\nx may not equal z, or you'd be dividing by zero again...\niii: x=z+m/s\nas part i, s may not be zero\n\nb) √(x²-4) may not be zero, so x may not be ± 2\nfurther, if -2<x<2, then √(x²-4) will be a complex number, so for both T and A to be real, x would have to be greater than 2 or less than -2...\n\nif the subject is changed to x:\nx=√[(A/T)²+4]\nthen T can't be zero
443911
I do not think you can.\nSo Bhaskara tried to solve the problem by writing n/0 = ∞. At first sight we might be tempted to believe that Bhaskara has it correct, but of course he does not. If this were true then 0 times ∞ must be equal to every number n, so all numbers are equal. The Indian mathematicians could not bring themselves to the point of admitting that one could not divide by zero. Bhaskara did correctly state other properties of zero, however, such as 02 = 0, and √0 = 0.
549551
The way this is written, there is nothing to solve because it's not an equation.\n\nTo use the zero product property, you want to factor an expression that equals zero. Then you conclude that if a product equals zero, one of its factors must be zero. To factor the expression in your problem, combine any like terms to simplify, then factor the simplified expression.\n\nHope that helps.
1164377
There are zeros at 2, -3/2 and -6, so\n\nf(x) = (x-2)*(x+3/2)*(x+6)\n\nf(x) = x^3 + 11/2*x^2 - 6*x -18
793079
The term with the highest exponent is x^5.  The coefficient of x^5 is one.\nThe constant term is 18.\nThe possible rational zeros are the factors of the constant term divided by each of the factors of the coefficient of the highest degree term.\nSince the coefficient of the term with the highest exponent is 1, that means the possible rational zeros are just the factors of the constant term.\nThe factors of 18 are 1, 2, 3, 6, 9 and 18.\nThe possible rational zeros could be negative or positive.\nAnswer: -1, +1, -2, +2, -3, +3, -6, +6, -9, +9, -18, +18
863763
D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
973592
You first look at the magnitudes of the terms when x is 1.  They are 4, 4, 7 and 2.  There is no way we can juggle their signs to make them add to zero.\n\nNext you look at the magnitudes when x is 2.  They are 32, 16, 14, and 2.  Yippee!  If one of the choices of sign can give us -32 + 16 + 14 +2, we have a root.  Yes, x = -2 makes the terms add to zero, so (x + 2) is a factor.\n\nBy synthetic division, 4x^3 + 4x^2 - 7x + 2 = (x + 2) (4x^2 - 4x + 1), and the quadratic factors as (2x - 1)^2.  I don't know why everybody else got it wrong.
638676
1.)\nstep one: if they are zeroes, use opp. \nexample....\n (x-2) (x+9) (x-7)          if you solve for x it should be opposite of given zeroes.\n\nstep two: FOIL the first two...\nyou'll get   x^ +9x -2x -18 which simplifies to\n\n(x^2 + 7x -18) (x-7) \n\nstep three: FOIL these two together...\nyou'll get     x^3 +7x^2 - 18x -7x^2 -49x + 126\n\nwhich simplifies to  \n   x^3 - 67x +126\n\n\n2.) I don't know about the second one... I could solve it if I spent some time, but start with...\n (x - 4i) (x+5) (x + 2i)\nthan just do what I did in the problem above, factor the first two, than factor your answer with the last one... and remember that i^2 is equal to 1...
879179
There is no limit.\n\nYou can see this by approaching from two different directions, along the x-axis and along the y-axis.If x is zero and y->0, then the formula becomes:\n\n   (siny)^2/y^2\n\nSince sin y/y -> 1 as y->0, this also approaches 1 as y approaches zero.\n\nOn the other hand, if y=0, and x->0, then the formula becomes:\n\n    x^2/(2x^2)\n\nwhich has a limit of 1/2 as x->0.
3037
This is not presented as an equation, so the goal appears to be to combine the rational expression --> x+10 \nThat's all you can do with the question as it is written.
1130519
Seperate the constants - you get:\n\n2x = -3\n\nDivide both sides by 2\n\n\nx = -3/2\n\ntake the cube root of -3/2\n\nIts an irrational about -1.2599
707641
another zero would be the conjugate of the one you know,\nthen try the rational zeros test, or even syn div
1003229
this is 0 dude...you will get x-x term which is zero and hence entire thing is 0.
1167006
If the zeros are: -4, 0, and 2, our factored equation must look like this:\nf(x) = x(x+4)(x-2)\n\nFoil it out:\n\nf(x) = x(x2 + 4x - 2x - 8)\nf(x) = x(x2 + 2x -8)\n\nf(x) = x3 + 2x2 -8x\n\nAnswer number 3.\n\nHope that helps.