DISCUSS DISCOVER: How Many Real Zeros Can a Polynomial Have? Give examples of po

Step 1
(a). Let a 3rd polynomial function P(x).
If the leading term is positive the end behavior is :
${\displaystyle y\to \mathrm{\infty }}$ as ${\displaystyle x\to -\mathrm{\infty }}$ and y ${\displaystyle y\to +\mathrm{\infty }}$ as ${\displaystyle x\to +\mathrm{\infty }}$
If the leading term is negative the end behavior is :
${\displaystyle y\to +\mathrm{\infty }}$ as ${\displaystyle x\to -\mathrm{\infty }}$ and ${\displaystyle y\to -\mathrm{\infty }}$ as ${\displaystyle x\to +\mathrm{\infty }}$
As we see in each case the end behavior is different as ${\displaystyle x\to -\mathrm{\infty }}$ and as ${\displaystyle x\to +\mathrm{\infty }}$ which means that the graph of a third degree polynomial must cross the x-axis at least once.
This means that every polynomial P(x) of degree 3 has at least one real zero.
Step 2
(b). Let a 4th degree polynomial P(x) .
${\displaystyle P\left(x\right)=x^4+x^2+3}$
The terms ${\displaystyle x^4}$ and ${\displaystyle x^2}$ are non negative as even powers of x and the constant term is positive.
P(x) is strictly positive. So the polynomial P(x) has no real zeros.
(c). Let the 3rd degree polynomial ${\displaystyle P\left(x\right)=(x-\sqrt{2})(x-\pi )(x-1)}$
Step 3
${\displaystyle P\left(x\right)=(x^2-x\pi -\sqrt{2}x-\sqrt{2}\pi )(x-1)}$
${\displaystyle =x^3-x^2-\pi x^2+\pi x-\sqrt{2}{x}^2+\sqrt{2}x+\sqrt{2}\pi x-\sqrt{2}\pi }$
${\displaystyle =x^3-(1+\pi +\sqrt{2})x^2+(\pi +\sqrt{2}+\sqrt{2}\pi )x-\sqrt{2}\pi }$
A 3rd degree polynomial with one rational zero ${\displaystyle x=1}$ and twp irrational zeros ${\displaystyle x=\sqrt{2},\pi }$ By our guidelines, we are supposed to answer first three questions only.