Question

state the positive number of positive real zero, negative real zero,imaginary zero of each funtion
1) r(x) =x^5-x^3-x+1
2) 5x^3+8x^2-4x+3
I attach lession

Answer 1

how I know how many real root or imaginary? after count the sign change?

Answer 2

You find positive reals by counting p(x) sign changes (then it's this number minus a multiple of two) then do the same for p(-x) to find negative reals (it's this number minus a multiple of two). Then all the other roots (the number of roots is the same as the degree) are imaginary.

So for your examples: 1) +ve reals either 0 or 2. -ve reals 1, and imag. either 2 or 4.
2) +ve reals either 0 or 2, -ve reals 1, and either 0 or 2 imaginary.

Answer 3

how u know ?

Answer 4

That's Descartes' rule of signs. Which bit is confusing you?

Answer 5

yes

Answer 6

1) positive sign; + - - +
yes no yes
negative sign: - + + +
yes no no
How I know number real zero,imaginary zero ?
help please!

Answer 7

Okay, so that's right. So you count the 'yeses' on the positive sign, 2. Then Descartes' rule of sign's says that it's this number or minus a multiple of 2 (it needs to still be positive though). So the number of positive roots is either 2 or 2-2=0.

Then same for negative roots. Number of yes=1. But as 1-2<0 there has to be 1 negative root (i.e. the number of yeses).

Now for imaginary roots it's all the ones left. So for Q1) the degree (highest power of x) is 5. So we know that the number of positive roots add the number of negative roots add the number of imaginary roots is =5. So we have the number of imaginary roots is (total roots)-(+ve roots)-(-ve roots) so here we have 5-2-1=2. OR (since +ve roots was 2 or 0) 5-0-1=4.

Answer 8

thank , take me so long can figout, easy to count sign change ,but sitll don't know zero roots , if u still aound can u help me more, I will post the same question