In descartes rule of signs, what do i do if the number of positive real zeros is 3, and then the number of?
551460
Well, you have to do one for each case.\n3 pos. and 2 neg. would leave 0 imaginary\n3 pos. and 0 neg. would leave 2 imaginary.\n1 pos. and 2 neg. -- 2 imaginary\n1 pos. and 0 neg. -- 4 imaginary\nAll the possible case have been covered.
750535
if f(x) change of sign twice\n\nso number of positive roots 0 or 2 based on Descartes rule\n\nfrom the given case 2 positive so 1 negative
1052440
The trivial zeros are at (-2K+0i), where K is a positive integer. The nontrivial zeros are all the others. It has been shown that the nontrivial zeroes must have a real part between 0 and 1.
339712
If 3 is a zero then P(3) = 0.\nP(3) = 27 + 36 - 21 - 10 = 32 \nThus 3 is not a zero.\nSince 3 is not a zero, then x-3 cannot be a factor.\nPossible rational factors are the positive and negative factors of 10\n1,2,5,10\nP(5) = 180\nP(10) = 1320\nThere are no rational factors greater than 3.\nAlso, there is not a zero between 3 and 4.\nC.\nThe zeros are -5, -1, and 2\nP(x) = (x+5)(x+1)(x-2)
665272
To build on what "Thin Kaboudit" was saying, you can take the limit of a function f(x) = 3/x as x goes to zero.\n\nLets say you decided to approach x from the positive side, slowly decreasing its value toward zero.\nThe closer and closer x gets to zero, the closer the function's value gets to [positive] infinity.\nNow lets say you decide to approach x from the negative direction, starting out with negative numbers and slow increasing them toward zero.  As you get closer and closet to zero from the negagtive side, the function goes toward negative infinity.\n\nSince the limit from the positive side does not equal the limit from the negative side (infinity does not equal negative infinity), the limit of the function does not exist since the function is not defined when x = 0.
842321
Absolute values can be very tricky so be careful with it. \nAll you need to know is that \n   *  if x<0  then |x|= -x\n   *  if x=0 then |x| = 0\n   * if x>0 then |x|=x\nIn your case, start with what's inside the absolute value.\nSo if 3-y>0 then x=3-y\nIn other words for y<3, y=3-x\nif 3-y=0, then x=0 (so for x=0 y=3)\nif 3-y<0 then x=-(3-y) = y-3\nThat is, for y>3, y=x+3\nSo you have to graph both functions y=3-x and y=x+3. When y<3 you take the first function and when y>3 you take the second one.\nFor your second function you forgot something but I'm guessing it was y=|x|. If it's the case just graph y=x (for x>0) and y=-x (for x<0). This will give you a symmetrical graph.\nIf you still have problems don't be scared to ask.\nHope this helps
950494
The rational zero theorem says that all possible rational zeros of a polynomial with integer coefficients, can be obtained by dividing the factors of the constant (in this case, it is -3) by the factors of the integer that is multiplied by the highest power of x (in this case, it it 2).\nAll factors of -3 are -1, +1, -3, +3.\nAll factors of 2 are -1, +1, -2, +2.\n\nThus, all possibilities are given by :\n-1/-1, -1/+1, -1/-2, -1/+2,\n+1/-1, +1/+1, +1/-2, +1/+2,\n-3/-1, -3/+1, -3/-2, -3/+2,\n+3/-1, +3/+1,+3/-2, +3/+2\n\nwhich leads to ±1, ±3, ±1/2, ±3/2 as the 8 possibilities.
914943
For the first one, the zero exponent should not be disregarded.  2x^0 is just one so it just eliminates that second 2x.  For the cubed part, the three needs to be distributed to the 2 and the x.  Is the negative one only on the x or is it on both the 2 and the x?  If it is only on the x, then your answer will be (2^3)(x^-3)=8x^-3 or 8/x^3.  If it is on both then you have (2^-3)(x^-3) which is 1/(8x).\n\nFor the second one, first flip the fraction, you get (4/3)^2 and then distribute the power to both the top and the bottom, youget (16/9).\n\nTHE NEGATIVE IN THE EXPONENT HAS NOTHING TO DO WITH A NEGATIVE IN THE BASE.  So just because there is a negative power, DO NOT put a minus sign in the base.\n\nFor the third one, just solve for y.\ny+2=3/8(x-16)\nget rid of the parenthesis and we have\ny+2=3/8x-6\nand then subtract 2 from both sides\ny=3/8x-2
860209
others have mentioned the values and let me explain the method\nf(x) changes sign 3 times so 1 or 3 positive root\nf(-x) = -x^3-8x^2 -19x -12 no change of sign so 0 negative root\n\nif it has a real 0 then it has to be factor of -12 that is 1/-1/2/-2/3/-3/6/-6/12/-12/4/-4\nwe need to test only positive roots 1 2 3 6 12 or 4\ntrying x = 1 we get 1-8+19 - 12 =0\n\nso (x-1) is a factor\ndy division\nx^3-8x^2+19x-12 \n= x^3-x^2-7x^2+7x + 12 x - 12\n= x^2(x-1) - 7x(x-1) + 12(x-1)\n= (x-1)(x^2-7x+12)\nquadratic we have to factor 12 so that sum is -7 we get -4 - 3\n\nso result = (x-1)(x-3)(x-4)
722679
60\n\nThe Babylonian sexagesimal number system, all Pythagorean triples have the property that the product of the 3 sides is a multiple of 60. I lived at a house which was no 60 in its street. And I am now 60.
417413
3 Zero's make thousand- 1000, 2000, unless it is like 10, 000 which has 4 or 100000 which has 5
817396
Since B is in the middle, lets set B as 0 for example purposes.  A would be any negative number (A<B) and C would be any positive number (B<C).  Since there are no 'or equal to' symbols, A<C.  This is true for all possilbe cases.
518111
As x->0, -3x->0. \nNow let -3x = z.\nThus the limit becomes\nLim z->0 [-3*{ln(1+z)}/z]\n= (-3)*Lim z->0 [{ln(1+z)}/z]\n= (-3)*1\n= -3\n\nSo 'c' is correct.
228129
It's an expression (i.e. not an equation (no equal sign).\n\nIf it was an equation that equaled zero:\n\nt^0 equals 1 so that quantity equals 2.\n\n18t^2-3=0\n\nt = +or- 1/sqrt(6)
835878
There's a better way of looking at it.\n\nFor every real number x in [0,1) you define D(x, n) as the nth decimal of x.  This is defined for every natural number n.\n\nFor x=1/3, we have D(1/3, n) = 3 for every n.  The number 0.3333... (the dots are important, they indicate that the nth decimal is a 3 for every n) is not "very close" to 1/3, it does not "tend toward" 1/3.  It is EXACTLY the number 1/3.  Show me a number other than 1/3 and I will show you that its decimal representation is not 0.33333...\n\nFor 0, we have D(0,n) = 0 for every n.  There is never a '1' decimal.  If you give me an arbitrarily small number 0.00000...1, the digit 1 is at a finite position p.  I can then reply with a number such that D(x,n)=0 for all n<=p, and D(x,p+1) = 1.  That number is greater than zero and smaller than the number that you gave me.  \n\nThe 0.5 = 0.49999... thing is just a quirk of decimal notation.  We can formally prove the equality as follows:\n\nLet a = 0.5, i.e. D(a, 1) = 5 and D(a,n) = 0 for all n>2.\nLet b = 0.4999..., i.e. D(b, 1) = 4 and D(b,n) = 9 for all n>2.\n\nProof by contradiction that a=b: Let's suppose on the contrary that a > b.  This means there must be a number x between a and b.\n\nLet's figure out the digits of x.  Since x is less that a=0.5, its first decimal must be a 4.  D(x,1) = 4.\n\nThe second decimal of x must be 9 since x > b `> 0.49.\nThe third decimal of x must be 9 since x > b >0.499.\nAnd so on, the nth decimal of x must be a 9 because x > b > the number y such that D(y,1)=4, D(y,i)=9 for i=2 to n and D(y,i) = 0 for i>n.\n\nTherefore x has exactly the same digits as b=0.499999... (I just showed that for every n, the nth digit of x is the same as the nth digit of b), and so x=b.  Contradiction, ergo a = b.\n\nThat's the difference between 0.499999... vs. 0.5 and 0.0000...1 vs. 0.  In the latter case, it is possible to find a number that is between the two.
1032854
given the zeroes of y:\nif x = -4, 0 or 2, then y = 0.  Therefore it passes through\n(-4, 0)\n(0,0)\n(2,0)\n\nIn other words -4, 0 and 2 are x-intercepts of the polynomial.\n\nThe polynomial (3rd degree) can be expressed as:\ny = x³ + ax² + bx + c\nIt can also be expressed as\ny = (x - x1)(x - x2)(x - x3) \nwhere n is a constant.\n\nWhen you substitute x1 to the expression\ny = (x1 - x1)(x1 - x2)(x1 - x3)\ny = (0)(x1 - x2)(x1 - x3)\ny = 0\n\nTherefore, x1 is a zero\n\nThe same goes with x2 and x3. Therefore you can put the zeroes into our expression\n\ny = (x - (-4))(x - 0)(x - 2)\ny = (x + 4)(x)(x - 2)\ny = x(x + 4)(x - 2)\nor\ny = x(x² + 2x - 8)\ny = x³ + 2x² - 8x\n\nThat is your 3rd degree polynomial! Hope this helps^_^
232213
I'll do the fourth problem.\n\n-3x< -6  or x+5< -2\n\nfirst deal with one half- \n-3x< -6 (now divide by -3)\nx>2 (when you divide by a negative the sign is reversed)\n\nthen the second half- \nx+5<-2 (get x alone, so subtract five from each side)\nx< -7\n\nso you have:\n2 < x  or  x < -7\n\nto graph this there is a line with an arrow to the left of the first slash and another line to the right of the second slash\n\n<-----/-------0--/-->\n____-7______2