SOMEBODY PLEASE FUCKING HELP ME!!!!!
Question:
The number a is call a double zero (or a zero of multiplicty 2) of the polynomial function P iff:
P(x)=(x-a)^2*q(x) and q(x)~=0.
Prove that a is a double zero of P iff a is a zero of both P and P' and P"(a)~=0.
(HINT: recall the factor theorum; see section 1.2)
Answer:
If P(x)=(x-a)^2*q(x), where q(a)~=0, then P(a)=P'(a)=0 and P"(a)~=0. Now suppose that P(a)=P'(a)=0 and P"(a)~=0. Then P(x)=(x-a)*g(x) for some polynomial g. Since P'(x)=(x-a)*g'(x)+g(x) and P'(a)=0, it follows that g(a)=0. Therefore g(x)=(x-a)*q(x) for some polynomial q, and P(x)=(x-a)^2*q(x). Finally P"(a)~=0 implies q(a)~=0.
The number a is call a double zero (or a zero of multiplicty 2) of the polynomial function P iff:
P(x)=(x-a)^2*q(x) and q(x)~=0.
Prove that a is a double zero of P iff a is a zero of both P and P' and P"(a)~=0.
(HINT: recall the factor theorum; see section 1.2)
Answer:
If P(x)=(x-a)^2*q(x), where q(a)~=0, then P(a)=P'(a)=0 and P"(a)~=0. Now suppose that P(a)=P'(a)=0 and P"(a)~=0. Then P(x)=(x-a)*g(x) for some polynomial g. Since P'(x)=(x-a)*g'(x)+g(x) and P'(a)=0, it follows that g(a)=0. Therefore g(x)=(x-a)*q(x) for some polynomial q, and P(x)=(x-a)^2*q(x). Finally P"(a)~=0 implies q(a)~=0.