Question

Graph of a biquadrtic polynomial intersects x -axis max at​

Answer 1

Answer:

It is trivial that if a four-degree polynomial has four real solutions, then f′′(x)=0 has two real solutions. But the converse has a counterexample(f′′(x)=0 has two real solutions can mean that f′(x)=0 has three extreme values. But that doesn't mean f(x)=0 has four real solution)

Answer 2

Answer:

Let p(X),q(X)p(X),q(X) be polynomials of degrees degp=m,degq=ndeg⁡p=m,deg⁡q=n. Then the difference r(X)=p(X)−q(X)r(X)=p(X)−q(X) is a polynomial of degree at most max{m,n}max{m,n} (in fact degr=max{m,n}deg⁡r=max{m,n} if m≠nm≠n or the leading coefficients differ). If p=qp=q, the difference turns out to be the zero polynomial and hence the polynomials agree everywhere (of course), but that case is excluded. Since p≠qp≠q, we have r≠0r≠0 and the points of intersection are the roots of r(X)r(X), hence there are at most degrdeg⁡r many.

Step-by-step explanation:

Hope it helps you.