The Student Room Group

Polynomial Factorisation

Show that if (x-a)^2 is a factor of the polynomial p(x) then p'(a)=0.

It's a question on a Past paper, I'm really not sure to go about it; I know that p(a)=0, and that p(x)=(x-a)^2(q(x)) + r(x) but I really don't know where to go?
if (x-a)^2 is a factor of the polynomial p(x) then are you sure you can say p(x)=(x-a)^2(q(x)) + r(x)?
Reply 2
Avatar for String.
String.
OP
11
Original post by WarriorInAWig
if (x-a)^2 is a factor of the polynomial p(x) then are you sure you can say p(x)=(x-a)^2(q(x)) + r(x)?



Should it be..

p(x)= q(x)(x-a)^2?
yes. You could have even had p(x)=(x-a)^2(q(x)+r(x)) if we're working with your example but obviously q(x)+r(x) is another polynomial and p(x)=q(x)(x-a)^2 is much simpler to work with.
Reply 4
Avatar for atsruser
atsruser
11
Original post by String.
Should it be..

p(x)= q(x)(x-a)^2?


Yes. Now simply compute p′(x)p'(x) and find its value at aa. (You'll need the product rule)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you