Show that the given value(s) of c are zeroes of P(x), and find all other zeroes of P(x).?
825629
P(1/3)=3(1/3)^4-(1/3)^3-21(1/3)^2-11(1/3)+6\nP(1/3)=3(1/81)-(1/27)-21(1/9)-11(1/3)+6\nP(1/3)=(1/27)-(1/27)-(7/3)-(11/3)+6\nP(1/3)=0\n\nP(-2)=3(-2)^4-(-2)^3-21(-2)^2-11(-2)+6\nP(-2)=3(16)-(-8)-21(4)-11(-2)+6\nP(-2)=48+8-84+22+6\nP(-2)=0\n\nNow factor (x+2) and (x-1/3) out of the original equation (I used synthetic division). That leaves you with:\n3x^2-6x-9=0\n3(x^2-6x-9)=0\n3(x-3)(x+1)=0\n\nSo, the other two zeros are -1 and 3.
1076304
(1) 1.3\n(4) 4\n(5) Right\n(6) 0\n(7) 0\n(8)0\n(9)12
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
961741
Im not going to prove it completely but I'll send you on your way..\n\nLook at a cubic, ax^3 + bx^2 + cx + d \nwe want to know for what value of x is this cubic = 0.\nLet's say r is a root.\nSet up the synthetic division:\n\nr || ..... a ........... b .............. c .................. d\n........................ ar .......... ar^2 + br ......... ar^3 + br^2 + cr\n───────────── ────────── ──────────\n......,....a......... ar+b ..... ar^2 + br + c......|| ar^3 + br^2 + cr + d\n\nin synthetic division, we say that if this last term,\nar^3 + br^2 + cr + d\nequals zero, then r is a root.\nNotice that the term is just the polynomial evaluated at x=r\n\nAnother way to look at it is:\nf(x) = ax^3 + bx^2 + cx + d\n= x( ax^2 + bx + c ) + d\n= x( x (ax + b) + c ) + d\n= x( x ( x (a) + b ) + c ) + d\nor,\ntake a,\nmultiply that by x, add b\nmultiply that by x, add c\nmultiply that by x, add d\nif that total is 0, then x is a root.\nThis is just what you do in synthetic division.
891987
The first step is to apply the Rational Roots Test to the polynomial to get a list of values that might possibly be solutions to the polynomial equation. You can follow this up with Descartes' Rule of Signs, if you like, to narrow down which possible zeroes might be best to check. Of course, if you've got a graphing calculator, it's a good idea to do a quick graph, since x-intercepts of the graph are the same as zeroes of the equation. Seeing where the graph looks like it crosses the axis can quickly narrow down your list of possible zeroes.\n\nOnce you've found a value you want to test, you use synthetic division to see if you can get a zero remainder. If you get a zero remainder, you've not only found a zero, but you've also reduced your polynomial by one degree. Remember that synthetic division is, among other things, division, so checking if x = a is a solution is the same as dividing out the linear factor x – a. This means that you should not return to the original polynomial for your next computation (for finding the other zeroes); you should instead work with the output of the synthetic division. It's smaller, so it's easier to work with.\n\nYou should expect to get some complicated solutions (that is, solutions containing square roots or complex numbers, or both). You will find these zeroes by applying the Quadratic Formula for the final quadratic factor.\n\nFind all the zeroes of the following polynomial: 2x5 + 3x4 – 30x3 – 57x2 – 2x + 24 \nFirst, I'll apply the Rational Roots Test--\n\nWait. Actually, the first thing I'll do is check to see if x = 1 or x = –1 is a root, because these are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because you can just look at the powers and the numbers. When x = 1, the polynomial evaluates as 2 + 3 – 30 – 57 – 2 + 24 = –60, so x = 1 isn't a root. But when x = –1, I get –2 + 3 + 30 – 57 + 2 + 24 = 0, so x = –1  is a root, and I can do the synthetic division with x = –1 right now.\n\n\n\nThis leaves me with the smaller polynomial 2x4 + x3 – 31x2 – 26x + 24. (Since I've divided out the factor x + 1, I've reduced the degree of the polynomial by 1. That's how I know this is a degree-four polynomial.) Now I'll apply the Rational Roots Test to get a list of values to look at:
561865
A permutation. P(n,r)=(n!)/((n-r)!). C(n,r)=(n!)/(r!(n-r)!)\nor (P(n,r))/(r!). x/something is smaller than x itself,(unless there's negatives or decimals, but we don't have to worry about that because r is positive) so the permutation is biggger than the combination.
813994
the first one is (x+10)(x+8)\nand the second one is your second answer
461122
Since f(x) is a polynomial of degree two in x,\nLet f(x) = ax^2 + bx + c, where a,b,c are constants.\n\nLet's find f(x) first.\n\nf(0) = 1 implies, c = 1.\n\nf(1) = 3 implies, a + b + c = 3\nor  a + b = 2 , or b = 2 - a  ...(1)\n\nf(-1) = 5 implies, a - b + c = 5\nor a - b = 4 ...(2)\n\nSubstituiting (1) into (2), we get,\n\na - [ 2 - a  ] = 4\n2a = 6 or a = 3.\nThus, b = 2 - a = 2 - 3 = -1.\n\nHence, f(x) = 3x^2 - x + 1\n\nNote that by long division or by the identity below,\n\n3x^2 - x + 1 = 3(x + 1)^2 - 7x - 2\n             \nWe get,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - [7x + 2]/ [1 + x]^2   \n\nAnd that together with below,\n\n[7x + 2]/ [1 + x]^2  = 7 /(x + 1) - 5 /(x+1)^2\n\nGives,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - 7 /(x + 1) + 5 /(x+1)^2\n  \nNow, we have an "integratable form".\n\n∫ [f(x) / (1 + x)^2] dx\n\n= ∫ [3x^2 - x + 1] / (1 + x)^2] dx\n\n= ∫ [ 3 - 7 /(x + 1) + 5 /(x+1)^2 ] dx\n \n= 3x - 7 log(x + 1) - (5/[x + 1]) + C\n\nwhere C is any arbitary constant.\n\nNote: Sometimes log(x + 1) can also be written as ln (x + 1)\n\nRemark. It is generally accepted that log and ln are both in base "e", the Euler number. If its in a different base, then usually its said or stated. For example logarithm of x in base 2 can be written as log_2(x).\nJust symbols :)\n\nRegards
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
762587
use remainder theorem\nhttp://en.wikipedia.org/wiki/Remainder_theorem\n"When p(x) is divided by (x-a), the remainder is p(a)"\nUR QUESTION:\np(x) = x^3+3x+k\n(x-a) = (x-1)\np(a) = 6\nhence, p(1)=6\nso, (1)^3 + 3(1) + k = 6\nhence, 1+3+k = 6\nso, k = 2\n\nhence, k = 2
528695
x to the n + y to the n = z to the n where n is an integer? ... damn it's been so long I can't remember for sure. Some Brit solved the proof already.
770351
if it has a rational zero it is integer and factor of 90\n1,2,3,5,6,9,10,15,30,45,90 with + and - sign\n\nyu can test for P(a) to be zero.\n\nby trial P(2) = 16-8-92-6+90 = 0\n\nso (x-2) is a factor\nP(x) = x^4 -2x^3+x^3-2x^2-21x^2+42x-45x+90\n= x^3(x-2) + x^2(x-2)-21x(x-2) - 45(x-2)\n= (x-2)(x^3+x^2-21x-45)\n\nnow to factor x^3+x^2-21x-45\n\nf(-3) = -27+9+63-45 = 0\nso x+3 is a factor\n\nx^3+x^2-21x-45\n= x^3+3x^2-2x^2-6x-15x-45\n=x^2(x+3) -2x(x+3) - 15(x+3)\n= (x+3)(x^2-2x-15)\n= (x+3)(x-5)(x+3)\nso zeroes are -3,-3,5 and 2
979279
take 3p common\n\n3p(2p^2-3p-2)\n= 3p(2p^2-4p+p-2)\n= 3p(2p(p-2)+1(p-2))\n= 3p((p-2)(2p+1)
740142
You're going to want to show, on a number line, what values of x satisfy the inequality.  Because you know that (-8) and (2) are both zeros of the function, you only have to test three regions:\n1) x is below -8.\n2) x is above -8 and below 2.\n3) x is above 2.\nFor each region, if the test point gives the function a value greater than 0, shade the region, otherwise, do not shade it.\n\n(You will end up shading one region.)
92762
A = P + Prt\n\n==> P(1 + rt)\n\n==> P = A/(1 + rt)\n\n<done>
1051369
Given a+b+c=0\na+b=-c\nCubing both sides, we get\n(a+b)^3=(-c)^3\na^3+3a^2 b+3ab^2+b^3=-c^3\na^3+b^3+c^3=-3a^2 b-3ab^2=-3ab(a+b)=-3ab(-c)=3abc\nTherefore a^3+b^3+c^3=3abc
622612
Joel, you need some help.  This is one of the easiest questions I have ever heard!  You should know this, man!  You're gonna fail your test if you keep cheating on here and don't actually figure out how it works.  Good luck, brotha'\n\nL = -2 or L(x) = -2