Can Any One Answer this :)?

Question

(x-a)(x-b)(x-c)......................(x-n).......(x-z)=??

Answer 1

The answer is PURPLE

Answer 2

Are there letters after z?

Let's look at the coeficients before x^n (when n is the number of factors): It is always 1.
(x - a)(x - b) = x^2 - (a + b)x + c

The coeficient before x^(n - 1) is -(a + b + c + ...)

In genaral for each k between 0 and n the coeficient before
x^k is -(the sum of all products of (n-k) roots.

It is easily seen when n=1
Nothing to prove for "x=a"

Let's assume it is true for 0
Let P(X) = (x -a1)(x-a2)...(x - an), 0<= k <= n

have a look at the coefficient before x^k, it is minus the sum of all products of (n - k) roots.

Let's take Q(x) = P(x)[x - a(n+1)]

The coefficient before x^n is the coefficient of the coefficient before x^n in p(x) * the coefficien t of x in [x - (a(n + 1)] , 1.

The free term is -a1*a2*...*a(n) * a(n + 1)

May k be greater than zero:
The coefficient before x^k in Q(x) is the coefficient before x^(k - 1) in P(x) - a(n + 1)*(the coefficient before x^k in p(X)), which is tminus the sum of all products of [n - (k-1)] roots of p(x) - a(n - 1)*(the sum of all products of (n - k) roots of p(x)]

Now, the sum of all products of [n - (k-1)] of P(x) is the sum of all products of (n+1 - k) roots which are not a(n + 1) of Q(x),
and a(n - 1)*(the sum of all products of (n - k) roots of p(x)] is the sum of all products of (n + 1 - k) roots in which a(n+1) participate.
Sum up both and get that the coefficient before x^k in q(x), is minus the sum of all products of (n + 1 - k) roots.

Answer 3

Hint: one of your factors is (x-x)...

Answer 4

(x-a)(x-b)(x-c)=x(-a-b-c)