Question

how do you factor and solve
(x-a)(x-b)(x-c)(x-d)...(x-z)

Answer 1

It is already factored. Each term (x-a), (x-b), etc is the factor of a polynomial that you would get if you multiplied all the terms together. (It would be difficult to factor that polynomial!)

as for solving: you need an equation to solve. if the whole thing were = to zero, for example, you could solve it. You would say:
if I have (x-a)(x-b)...(x-z)=0
and x were a, then the first term (x-a) would be (a-a)=0 and 0 times anything is 0. so x=a would "solve" the equation: you would get 0=0 which is true.
you can also say x=b would solve the equation, and x=c and so on...

Answer 2

i'm sorry i meant how do you F.O.I.L the problem
but thank you so much for your response

Answer 3

Oh, in that case, the previous post (since deleted) was answering that question.
but after the first multiply FOIL (first, outer, inner, last) does not quite fit.

(x-a)(x-b)= x^2 - (a+b)x +ab
now multiply by (x-c):
(x^2 - (a+b)x +ab)(x-c)= x*(x^2 - (a+b)x +ab) + -c* (x^2 - (a+b)x +ab)
you distribute (x^2 - (a+b)x +ab) over x-c
now distribute the x and the -c and combine like terms:
x*(x^2 - (a+b)x +ab) + -c* (x^2 - (a+b)x +ab) becomes
x^3 -(a+b)x^2 +ab x -c x^2 + (a+b)c x - abc or
x^3 -(a+b+c)x^2 + (ab+ac+bc)x -abc
now multiply by (x-d)
it will get very complicated.

Answer 4

thank you so much that really helps