Question

State how many imaginary and real zeros the function has.
f(x) = x5 + 7x4 + 2x3 + 14x2 + x + 7

3 imaginary; 2 real

4 imaginary; 1 real

0 imaginary; 5 real

2 imaginary; 3 real

Answer 1

Counting sign changes of f(x) .......... 0 - There are NO Positive Real zeros.
Counting sign changes of f(-x) .......... 5 - There is one Negative Real zeros, and maybe 3 and maybe 5
There are only TWO suspects for Negative Rational zeros. x = -1 and x = -7. Youse the remainder theorem to check them out.

How are we doing?

Answer 2

Notice how we've already thrown out #1 and #3.

Answer 3

so would it be D

Answer 4

?

Answer 5

No, not quite. I switched words.

Real: 1, 3, or 5
Rational: x = -7 or x = -1 -- You ahve to try these.
I said nothing of possible negative irrational zeros.

Once you find x = -7 and reduce the polynomial, you will see the other four zeros without much trouble.

Answer 6

so.. itd be 4 imaginary; 1 real

Answer 7

Maybe. You have to prove it. In this case, it is relatively easy to find them since there is a rational zero and the reduced polynomial is trivial - this is not expected for a quartic polynomial.