Zero product property    w(w-4)=0?
600360
In w(w-4) there are two factors being multiplied.  One, as you know, is w-4.  The other is w.  To use the zero factor property, you set each factor equal to 0.  This means that you will set w equal to 0 as well as setting w-4 equal to 0.
782446
x:-    0** 2** 4\ny:-  -4** 0** 4\n\ni hope that this helps
134065
Your statement translates to the following algebraic equation.\na+b=a-b=a*b\nSolving a+b=a-b, we get b = 0; \nSubstituting b=0 and solving a+b=a*b, we get a = 0;\nSo the two numbers are zeroes.
242996
Technically the answer is infinite.\nThere are infinite numbers that would give you a product of four. \nBut the width can be [lim width->4] so the hight will approach 0. \nThe biggest width would be close to (approaches) 4 but not 4.
659301
Zero divided by any number (except zero) is zero.
965399
8(w+4)=w^3+2w^2-8w\n<=> w^3 + 2w^2 -16w - 32 = 0\n<=> w^2(w+2) -16(w+2) = 0\n<=> (w^2 - 16)*(w+2) = 0\n<=> (w-4)*(w+4)*(w+2) = 0
699032
wx - x =\n\nx(w - 1)\n\n- - - - - - -s-
707641
another zero would be the conjugate of the one you know,\nthen try the rational zeros test, or even syn div
339198
2x0 = 0\nIt'll be the same as today... :P
148878
1power0 = 1\n1power1 = 1\nso that 0=1
436432
X = (0*Y) -5\nY = 0 * X\n\nbut basically just write X = -5 and thats it
971462
48=2(20+w)\n24=20+w\nw=4
187658
it's defined \n\n0^0=0
556493
> does the 'irreducible' part imply that it's not 0 for any value in Zn?\n\nOnly for degree less than or equal to three.  It means the polynomial is not a product of polynomials of lesser degree.  A fourth degree polynomial which is the product of two irreducible second degree polynomials will have no zeroes in Zn but is still not irreducible itself.  But if the degree of a reducible polynomial is less than or equal to three then at least one factor is linear and thus there will be a zero in Zn.\n\n> does the 'primitive' part imply that the powers of x generate the entire field, F(n^2)\n\nAlmost: a degree k irreducible polynomial p(x) is primitive if x generates all of the units (nonzero elements) of F(n^k).\n\n> If i'm right, then showing a polynomial is irreducible in Z11 is going to be about as tedious as it gets...\n\nIf you do it by hand. :-)
552485
(4)(1)(-10)=\n(4)(1)=4(-10)=-40
589852
Sigma sign with k=0 on bottom and 4 on top; (-1)^k over (2k+1)^2
577215
This again is a direct theorem from the field axioms: http://mathworld.wolfram.com/FieldAxioms.html . \nSuppose a*b = 0. If a is non-zero, multiply both sides by a^(-1) to get a^(-1)ab = a^(-1)0, and thus b = 0, since the product of 0 and any number is 0.\nIf b is non-zero, multiply both sides by b^(-1) to get the similar result a = 0. Thus, either b or a must be 0. \nThe final case is that both a and b are 0, which is trivial.\nFurther discussion of this theorem can be found in the source.\nThis theorem can be used to solve equations by getting 0 on one side of the equation and factoring the other side completely, then using the theorem to imply that at least one of the factors must be zero. For example, x^2 = -3x - 2 can be rearranged to be x^2 + 3x + 2 = 0. We factor the left side to read (x + 1)(x + 2) = 0. From the theorem we know that either x + 1 = 0 or x + 2 = 0 or both. Thus we have two solutions, x = -1 or x = -2. It is obvious that in this case, only one factor could be 0.