thomas49th said:
P(x) =\left | \begin{array}{ccc} 1&1&1\\ x&b&c\\ x^2&b^2&c^2 \end{array}\right |
the determinant for the first one is:
P(x) =\left | \begin{array}{cc} b&c\\ b^2&c^2 \end{array}\right |
+(bc² - cb²)
That is the constant coefficient of the polynomial but let's not call it P(x). We have already called the top one P(x).
x is:
P(x) =\left | \begin{array}{cc} 1&1\\ b^2&c^2 \end{array}\right |
-(c²-b²)
Again, this is the coefficient of x, but don't call it P(x).
and x²:
P(x) =\left | \begin{array}{cc} 1&1\\ b&c \end{array}\right |
+(c-b)
That okay?
Yes, the coefficient of x
2 is (c-b). Again, not P(x) though.
If P(x) is a second degree polynominal then (x-b)(x-c)
Everything seem okay so far?
Thanks :)
Tom
Good. You are almost there. Since P(b) and P(c) are zero P(x) must have factors (x-b)(x-c). But that doesn't completely determine P(x) because P(x) might be a constant times that and it would still have those roots. So at this point you know:
P(x) = A(x-b)(x-c)
for some constant A. Notice that that constant A would be the coefficient of x
2 if you multiply it out. But you know from above that the coefficient of x
2 in P(x) is (c-b). So A must equal (c-b). This tells you that
P(x) = (c-b)(x-b)(x-c)
So now we have an identity:
P(x) =\left | \begin{array}{ccc} 1&1&1\\ x&b&c\\ x^2&b^2&c^2 \end{array}\right | = (c-b)(x-b)(x-c)
Since this is an identity it is true if we put x = a in both sides:
P(a) =\left | \begin{array}{ccc} 1&1&1\\ a&b&c\\ a^2&b^2&c^2 \end{array}\right | = (c-b)(a-b)(a-c)
That last determinant is what you started with and we have shown what its value is. If you multiply out the determinant the long way you might have difficulty showing it factors nicely like this. Like I said when we started, I wanted to show you a way to expand this determinant that you likely wouldn't think of. It's pretty nifty and I hope you like it.
