Can someone tell me what the sentence "if k > 1 then

>>8979471
It means the slope of P(x) at x=r1 is zero.

Let P(x)=[math] (x-r_1)^{k_1} (x-r_2)^{k_2} ... (x-r_n)^{k_n} [/math]
Then slope of P(x) is P'(x)=[math] (k_1)(x-r_1)^{k_1 -1} (x-r_2)^{k_2} ... (x-r_n)^{k_n} + (x-r_1)^{k_1} (k_2)(x-r_2)^{k_2 -1 } ... (x-r_n)^{k_n} +... (x-r_1)^{k_1} (x-r_2)^{k_2} ... (k_n)(x-r_n)^{k_n - 1} [/math]
So P'(r1) = (r1-r1)^(k1 - 1)(something not 0)+0+....+0. If k1=1 then (x-r1)^0 = 1 and the slope is not zero and not flat otherwise k1>1 and you have 0^k1-1 = 0 so the slope is zero.

>>8979505
Thanks, I understand it a little better now. I think I was a little confused because I forgot that regardless of the multiplicity, it's still a single point on the x-axis. and if k > 1 we know it's a slope of zero

>>8979471
I'll just post this here in case you don't intuitively understand why 1 and 2 hold.

P(x) can always be written as a constant times (x-r1)^m1 ....(x-rn)^mn times a bunch of terms that are of the type x^2+bx+c where b^-4c is less that 0.

Let's study the behaviour of the polynomial around r1.
We have that every term in the polynomial except (x-r1)^m1 is non 0 in a region around r1 e.g. (r1-ε,r1+ε). Those terms also hold their sign in that region, cause if they changed signs they have to become 0 at some point (since they are continuous).

If m1 is odd, then
for x in (r1-ε, r1) you'll have P(x)= negative times the terms that don't change sign
for x in (r1, r1+ε) you'll have P(x)= positive times the terms that don't change sign
From this it is obvious to see that in that region, P(x) Crosses the x axis, either from down to up, or from up to down.

If m1 is even, then
for x in (r1-ε, r1) you'll have P(x)= positive times the terms that don't change sign
for x in (r1, r1+ε) you'll have P(x)= positive times the terms that don't change sign
From this, you can see that P(x) stays either bellow the x axis or above it for all x in (r1-ε,r1+ε) except at r1 where it only touches it.