OpenStudy (anonymous):

State how many imaginary and real zeros the function has. f(x) = x4 - 8x3 + 17x2 - 8x +16

4 years ago
OpenStudy (anonymous):

0 imaginary; 4 real 2 imaginary; 2 real 3 imaginary; 1 real 4 imaginary; 0 real

4 years ago
OpenStudy (cwrw238):

you can apply rational root test here - though my memory of it is a bit vague i'm afraid as the last number is + 16 i would try x = 2 as 1 root

4 years ago
OpenStudy (cwrw238):

16 -64 + 68 - 16 + 16 = 0 yes x=2 is one root

4 years ago
OpenStudy (cwrw238):

i would venture to say that that might be a repeated root

4 years ago
OpenStudy (cwrw238):

because of the coefficients in the equation

4 years ago
OpenStudy (anonymous):

So 2 imaginary and 2 real?

4 years ago
OpenStudy (cwrw238):

maybe 4 real

4 years ago
OpenStudy (anonymous):

So four real and zero imaginary?

4 years ago
OpenStudy (cwrw238):

i'm not absolutely sure - I'd have to revise this stuff sorry i cant be of any more help

4 years ago
OpenStudy (anonymous):

Thanks anyway

4 years ago
OpenStudy (cwrw238):

i'm sure there are 2 real roots - both = 2 so its either option 1 or option 2

4 years ago
OpenStudy (cwrw238):

yw

4 years ago
OpenStudy (anonymous):

I'll go with the first one and see what happens. Thanks :)

4 years ago
OpenStudy (cwrw238):

I've looked at this again. I made an error - the root is not 2 but 4. applying Descartes rule of signs there is either 0,2 or 4 real positive roots and no real negative roots. But we have found one real positive root which is 4. I suspected that this is a double root so did a long division . If 4 is a double root then the polynomial is divisible by (x-4)^2 or x^2 - 8x + 16 - which it is and the quotient is x^2 + 1 which if you equate to zero gives 2 imaginary roots i and -i so the correct option is B, 2 imaginary and 2 real

4 years ago