p is obviously zero because any number x plus 0 equals itself.
x-a is a factor of the polynomial p(x),if p(a)=0.also,if x-a is a factor of p(x), p(a)=0.
There is no "you" in P = V x I
x2 + 3px + p = 0 x2 + p(3x + 1) = 0 p(3x+1) = -x2 p = -x2/(3x+1) So p can have any value at all. In fact, around x = -1/3, p goes asymptotically to + and - infinity.
111684
p is obviously zero because any number x plus 0 equals itself.
If p, q, r, ... are the roots of the equations, then (x-p), (x-q), (x-r), etc are the factors (and conversely).
p must equal zero p = 0 (7 x 0) + 8.4 = 8.4 0 + 8.4 = 8.4
x-a is a factor of the polynomial p(x),if p(a)=0.also,if x-a is a factor of p(x), p(a)=0.
p = -log[H+] = 12.4
There is no "you" in P = V x I
x2 + 3px + p = 0 x2 + p(3x + 1) = 0 p(3x+1) = -x2 p = -x2/(3x+1) So p can have any value at all. In fact, around x = -1/3, p goes asymptotically to + and - infinity.
P = Cx 'P' = the product 'C' = any integer
111684
5
5.8
p(x) = 0 => x4 + 2x3 + x2 + 8x - 12 = 0=> (x + 3)*(x - 1)*(x2 + 4) = 0So the imaginary roots of p(x) are the imaginary roots of x2 + 4 = 0that is x = ±2i