Does anyone know how to find the zeros of ploynomials?
958565
Well, for quadratics, which are polynomials of degree 2 (i.e. the largest term is ax^2), the quadratic formula always works.  Of course, it's scary looking if square roots intimidate you.
500255
This is called an inequality. First, find the zeros (in this case, -4 and 9). Next, make a number line:\n<------(-4)----------(9)----->\nNow, determine where inside and outside the zeros the inequality will be true or false. Finally, express your answer in terms of the inequality.\nExample: x+12 > 0\nAnswer: x > -12
96233
+5,+12, -5, -12, +3.3
775090
You have to find the two polynomials that when multiplied by each other will equal the trinomial.\n\nx2 + 8x - 9\n\n(x + 9) (x - 1)\n\nx*x + -1x + 9x -9\n\nx2 + 8x - 9
204914
I only need four 0.\nbinary:0!+0!+0!+0!=100
596283
add them up like this \n4+2+(-3)+6+5+(-3)+(-2)+0+(-1)+(-4)+3\ndivide the result by 11 because that is how many numbers there are in your list.  That will be your answer!
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)]
367158
(0! + 0! + 0! + 0! + 0!)!\n\n\n0^0 is undefined.
437003
let the numbers be \nx-d, x and x+d. \n\ntheir sum is zero => x = 0\n\nThe sum of their squares is 8 => 2*d^2 = 8 => d = 2\n\nHence the numbers are -2,0,2
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
959413
numbers ending in -0 or -5
793059
give values of -3,-2,-1,0,1,2,3 to x and find the corresponding values of y\nyou get the ordered pairs\nnow plot
156464
u can't devide by (a-b) as (a-b)=0
424740
Unfortunately there is not. There is however an algorithm that will allow you to work out square roots by hand.
887079
The long hand method for dividing polonomials is much like normal division\n\n\nput the denominator to the left of your square root looking symbol and then your numerator under your that symbol....\n\n\n1.take the first term from the denominator and muliply it by whatever you need to in order to get the first term in the numerator in this case that is 3x(keep this in some part[if you use normal division method it will be on top of symbol as it is part of your answer)\n\n2. now multiply the operand(3x in this case) by your entire denominator for this= 3x^4 + 3x^3 - 27x\n\n3. now take the numerator and subtract what you got from it\n(3x^4 - 6x^3 - 14x^2 - 6x + 18) - (3x^4 + 3x^3 - 27x) = -9x^3 -14x^2 + 21X + 18\n\n4. now repeat steps 1-3, with your product from #3(every time) as your new numerator, untill you need a negative exponent to complete step 1\n\nyour overall answer is 3x - 9 - {(5x^2 + 21x - 63)/(x^3 + x^2 -9)}\n\nsorry my steps are so confusing but I rarely help others... so I came here to kind of practice for tutoring
875039
The first five are simple\nP(5,1) + P(5,2) + P(5,3) + P(5,4) + P(5,5) \n5 + 20 + 60 + 120 + 120 = 325\n\nSix and seven are a little more complicated\n6!/2! + 7!/2!2!\n360 + 1260 = 1620\n\nEight is special because you limit the maximum.\n\nList of bad values\n9x,xxx,xxx\n7x,xxx,xxx\n69,xxx,xxx\n67,xxx,xxx\n66,xxx,xxx\n65,9xx,xxx\n65,7xx,xxx\n65,6xx,xxx\n\n8!/3!2! - (2 * 7!/3!2!) - (3 * 6!/3!) - (3 * 5!/3!)\n3360 - 840 - 360 - 60 = 2100\n\nNow add the results.\n325 + 1620 + 2100 = 4045
601075
check out this website:\n\nhttp://www.wehirealiens.com/resultsall.asp\n\nI'll boycott with you. Anyone who says one or two people can't make a difference is wrong. It starts with one, then two, then ten, then a hundred....keep adding zeros and it starts to add up.