Find a ploynomial of the specified degree that has the given zeroes?
965701
f(x) = x(x+2)(x-2)(x-4)\n\nMultiply it out if you need it in general form.\n\nThere could also be a constant in front of the x.
378781
-1*100\n-2*50\n-4*25\n-5*20\n-10*10\n~or~\n1*-100\n2*-50\n4*-25\n5*-20\n10*-10
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
540459
something close to zero
833237
so far, so good\n\nn!/(n-4)! = 56 [n!/(n-2)!] \n\nmultiply both sides by (n-2)!/n!\n\n[n!(n-2)!] / [n!(n-4)!] = 56\n\n(n-2)! / (n-4)! = 56\n\n(n-2)(n-3) = 56\n\nYou could multiply this out and solve the quadratic, in which case you will get one positive and one negative root, or if you remember that 56 = 8*7, then that quickly leads to n = 10
961234
Let x = y^(1/3)\nx^2 - 9x + 8 = 0\n(x - 8)(x - 1) = 0\nx = 8 or 1\ny^(1/3) = 8 or 1\ny = 2 or 1
629310
nything that doesn't involve either zero or U
465556
Given that f(x) is a polynomial of degree EIGHT such that\n\nf(m) = 1/m for m = 1,2,3,4,5,...,9\n\nFind f(10)  \n\n\n*Answer: 2/10\n\n*Solution:\n\nConsider another polynomial, g, defined as g(x) = x f(x) - 1\n\ng(1)=0, g(2)=0, …, g(9)=0\n\nSo g(x) = A(x-1)(x-2)…(x-9)\n\ng(0) = 0 f(0) - 1 = -1\n\nalso, g(0) = A(-1)(-2)…(-9), so A = 1/9!\n\ng(10) = A(9)(8)…(1) = (1/9!)(9!) = 1\n\ng(10) = 1 = 10 f(10) - 1, so f(10) = 2/10\n\n\n**Or try doing some research on "Rabin's method"
773850
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.
547629
assuming u is an unknown variable y=0
762440
Just split it into parts - when is x-x^2<0, and when is it >0?\nThat bit has roots 0 and 1, so it is positive between 0 and 1, negative everywhere else.\n\nThus, ∫(x=-1 to x=0) (x^2-x) dx + ∫(x=0 to x=1) (x-x^2) dx + ∫(x=1 to x=3) (x^2 - x) dx\n\n= [(1/3 x^3 - 1/2 x^2) from -1 to 0] + [(1/2 x^2 - 1/3 x^3) from 0 to 1] + [(1/3 x^3 - 1/2 x^2) from 1 to 3]\n= (0- -5/6) + (1/6 - 0) + (9/2 - -1/6)\n= 17/3.
1146457
vertex (h,k)\nfocus (h,k + p)\n\n(x - h)^2 = 4p(y - k)\n\nvertex : (2,-6)\nfocus : (2,-6 + p)\n\nk + p = -8\n-6 + p = -8\np = -2\n\n(x - h)^2 = 4p(y - k)\n(x - 2)^2 = 4 * -2(y - (-6))\nx^2 - 2x - 2x + 4 = 4 * -2(y + 6)\nx^2 - 4x + 4 = -8(y + 6)\nx^2 - 4x + 4 = -8y - 48\nx^2 - 4x + 8y + 52 = 0\n\nANS : d.) x^2 - 4x + 8y + 52 = 0
332120
Computer Science: 2B || !2B\n\nLogic: 2B V ~2B\n\nAlgebra: x² - 4B² = 0\n\nAs this factorises by difference of two squares to\n(x - 2B)(x + 2B) = 0\nGiving two possible solutions\nx = 2B, -2B
867199
domain {3.1,0}\nrange={-6,8,-9}
1076304
(1) 1.3\n(4) 4\n(5) Right\n(6) 0\n(7) 0\n(8)0\n(9)12
367158
(0! + 0! + 0! + 0! + 0!)!\n\n\n0^0 is undefined.
751318
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 :)