If the leading coefficient of a polynomial is $x^{3}$, does it mean that the graph would always intersect the $x$ axis at $3$ points?

Question

I had learnt that this was the case. Or is it not like that? Does the power of a polynomial indicate the number of times the graph cuts at the x axis?

Answer 1

Try $p(x)=x^3$. ${\ \ \ \ \ \ \ \ }$

Answer 2

No. Take $f(x)=-x^3+3x^2+7$ for example.

Answer 3

No. But the degree of the polynomial gives you the maximum number of zeros. The reason is that if $x=a$ is a zero, i.e., if $p(a)=0$, then you can write $p(x)=(x-a)\cdot q(x)$, where the degree of $q(x)$ is one lower than that of $p(x)$. Do this $n$ times, where $n$ is the degree of $p(x)$, and you end up with a constant factor at the end.

Over the complex numbers, as opposed to the reals, a polynomial will have as many zeros as its degree, if you count them with /multiplicities/. More precisely, you can write a complex polynomial as $p(x)=c\cdot(x-a_1)\cdots(x-a_n)$, and the multiplicity of a zero is just the number of times the zero equals some $a_j$.

Answer 4

Generally speaking 3 is the maximum number of distinguished intersection with the real axis. Moreover the fundamental theorem grants, but in complex field, the existence of "three" roots yet their multiplicity may be more than 1. This is the case for $x^3 = 0$ were $0$ is a root with multiplicity 3. And intersection with real or complex axis too have multiplicity 3. The intuitive approach to the concept of multiplicity is this: a little perturbation of the equation will separate three distinct roots near the original one. For example: $x^3-0.1x$ has three distinct real root. Consider anyway the equation $x^3+x=0$ here you will have just one real root and two complex $\pm i$ roots and there is no little perturbation making the equation have three real roots.