Question

ALGEBRA 2 help please;Question in the attached

Answer 1

Answer:

$f(x) = x^{5} -3x^{2} -2$ has 5 real and 0 non-real zeroes.

Step-by-step explanation:

Here, the given polynomial is $f(x) = x^{5} -3x^{2} -2$

Now, by FUNDAMENTAL THEOREM OF ALGEBRA:

A polynomial of  degree n can have at most n roots.

So, here the number of roots f(x) can have = 5

Now, examine the change in the sign of f(x) and f(-x)

$f(x) = x^{5} -3x^{2} -2$

Signs of f(x) is  + - -.

So for f(x)  the sign changes only once.

Now,  $f(-x) = (-x)^{5} -3(-x)^{2} -2$

or,  $f(-x) = -x^{5} -3x^{2} -2$

Here, signs are - - -

So,for f(-x)   the SIGNS DO NOT CHANGE.

So it has no negative real zero.

Hence, $f(x) = x^{5} -3x^{2} -2$ has 5 real and 0 non-real zeroes.