What is the greatest number of possible rational zeros of this polynomial:\n\nP(x) = 2x^4 + 10x^3 -3x^2 -8x +4
217039
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.
460531
This would be a special case of the Fermat-Catalan Conjecture.\n\nThis conjecture is that there are only a finite number of solutions of x^p + y^q = z^r for 1/p + 1/q + 1/r < 1.  Ten solutions are known, but all of them have the three exponents different (except for 1^p + 2^3 = 3^2) and one exponent equal to 2.\n\nYour conjecture would be the special case that there are no solutions with p > 2 and q = r.  It is already known that there are no solutions with p = q and r > 2 (Darmon and Merel, 1997).
397124
Tip:  If the sum of the coefficients of the polynom are zero, then x =1 is a zero of the polynom.\n\nx = 1\nIf you need more roots do the synthetic divisiona and then use the quadratic formula.
589182
Here is my proof. Suppose P(x) is a polynomial in odd degree, for example 4x^5 + 3x^4 - 5x^3 + 2x^2 - 6x + 9. For large enough values of x, the term in x^5 will be so much larger than any of the other terms that it will swamp the smaller powers of x, and x^5 alone will determine the value of P. \n\nNow, if P is of odd degree and x is very large and positive, P will be positive; if x is very large and negative, then P will be negative. Therefore, somewhere between these two values of P there must be at least one point where P = 0. QED.
665999
Domain lists all possible x values. A normal polynomial function has a domain of all real numbers, because EVERY x value works.\n\nYour task is to find x values that DON'T work...and since you cannot divide by zero, then any x in the denominator that makes x^4+1 equal to zero will not be part of the domain.\n\nSo, what do you know about x^4+1?\n\nIt is an even power, so even if you put a negative number in for x, x^4+1 will be positive (since adding one will only make it more positive). If x=0, that is still positive, because of the plus one.\n\nx^4+1 is ALWAYS positive for any x...therefore it is NEVER equal to zero. Therefore any x value works, and your answer is all real numbers!!
1391
Here's the best I can do. The answer I got is:\n\n24x^4 - 10x^3 + 690x^2 + 3190x = 1217
364643
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)
463871
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.
349652
(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=================================
479617
The zeros tell us that x-1, x-2 and x-4 are factors, so the polynomial is  \nf(x) = k*(x-1)(x-2)(x-4) where k is some constant.\n\nf(0) = k*(-1)(-2)(-4) so you just choose the value of k that makes this -16.\n\nIf 3 + i is a root then so is 3 - i, and multiplying\n(x - (3+i))(x - (3-i)) gives\n\nx^2 - 6x + 10, so this is your quadratic polynomial.\n\nObviously the linear factor is x - 4 since  4 is a root.  Leading coeff 1 means you don't have to bother to multiply these factors by some other constant coefficient.
500894
are you talking about a polynomial :\na + bx +cx^2 + dx^3 +...\n\nbecause those a,b,c,d... do not appear any where on the graph... you have to look at the intercepts and the extreama, and the inflection points to get any good idea about the shape of the graph
210670
Do you mean "zero of a polynomial?" If so, then read on...\n\nA zero of a polynomial is the value or values of the variable (usually x) that cause the polynomial to equal zero.\n\nExample 1: The zero of the monomial 3*x is 0 because 3*0 is zero.\n\nExample 2: The binomial x + 2 has it's zero at x = -2 because if you substitute -2 in place of x, then you get \n---> -2 + 2 = 0.\nTherefore, -2 is the zero of the given binomial.\n\nExample 3: Find the zeros of the trinomial x^2 - x - 6. That is, solve x^2 - x - 6 = 0. If you factor the trinomial, then you get \n---> (x + 2)(x - 3) = 0. \nIn this problem there are two possibilities. Either \n---> x + 2 = 0  or  x - 3 = 0. \nSolving each gives x = -2 or x = 3. Therefore, the zeros of the given trinomial are -2 and 3. \n\nIf you substitute -2 into the given trinomial, then you have the following:\nx^2 - x - 6 \n= (-2)^2 - (-2) - 6 \n= 4 + 2 - 6\n= 6 - 6\n= 0 \n\nIf you substitute 3 into the given trinomial, then you have the following:\nx^2 - x - 6 \n= (3)^2 - (3) - 6 \n= 9 - 3 - 6\n= 6 - 6\n= 0 \n\nThere are no other values of x that will cause our given trinomial to equal zero.\n\nDefinitions\nA monomial is a polynomial with one term.\nA binomial is a polynomial with two terms.\nA trinomial is a polynomial with three terms.\nA polynomial with more than three terms is just called a polynomial.\n\nNote: If you graph a polynomial, then the graph will touch or cross the x-axis at the zeros of the polynomial. Zeros of a polynomial are also called solutions, roots, and, in some cases, x-intercepts.
376210
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
380222
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
158178
Did you mean to have the yx^2-30x+8 as your polynomial (a polynomial in two variables?)\nFor y=7   or 7x^2-30x+8\n\n7x^2-30x+8 = 7x^2-28x-2x+8\n\n7x^2-28x-2x+8= 7x(x-4)-2(x-4)\n\n7x(x-4)-2(x-4)= (7x-2)(x-4)
96179
x^4 + 4\n\nWell, you can add 0 in the form of +a - a, so\nx^4 + 4 +4x^2 - 4x^2\n\nRegrouping them, we get\nx^4 + 4x^2 + 4 - 4x^2\n\nUsing factoring, we get\n(x^2 + 2)^2 - (2x)^2\n\nThen,\n[x^2 + 2 + 2x][x^2 + 2 - 2x]\n\nRearranging them for a more beautiful sight, we get\n{x^2 + 2x + 2)(x^2 - 2x + 2)\n\nThanks For the Problem ^_^
345302
so the Q is basically asking for two binomials which, if multiplied will end up as a polinomial...\nAs a matter of fact any two binomials can form a polynomial.\n(2X+3) (x-7)...this is an e.g.