Question

Find the number of possible positive and negative real zeros of:
1. f(x)=x^4-x^3+2x^2+x-5
2. f(x)=3x^3-4x^2-5x+1
3. f(x)=5x^6+4x^5+3x^4-2x^3+2x^2-x-1

Answer 1

well let us use the Descates Rule of Signs

f(x)= x^4 - x^3 + 2x^2 + x - 5
+ - + + -
yes yes no yes

so the sign changed 3 times, that means
3 or (3-2) real positive roots
---------------------------------------
now let's check for f( -x )
f( -x )= x^4 + x^3 + 2x^2 - x - 5
no no yes no

so the sign changed only once for f( -x )
meaning that, there's only 1 negative real root

Answer 2

so that equation is a 4th degree polynomial, thus 4 roots

it either has 3 real positive ones an 1 negative
or
1 positive one and 1 negative, leaving room for 2 complex ones, for a total of 4

Answer 3

Would the second one be 2 pos, 1 neg then?

Answer 4

And the last one 4 pos, 2 neg?

Answer 5

@jdoe0001

Answer 6

the 2nd one... let's see

f(x)= 3x^3 - 4x^2 - 5x + 1
yes no yes

2 OR (2-2) real positive ones

f( -x ) = -3x^3 - 4x^2 + 5x + 1
no yes no

1 real negative only

yes 2 real positive and 1 negative # is a 3rd degree so 3 roots
OR
(2-2), 0 real positive ones and 1 negative and 2 complex ones

Answer 7

ok perfect! and the third? @jdoe0001

Answer 8

how many signs change did you get for f(x) and f ( -x ) ?

Answer 9

I got 4 positive and 2 negative as my answer. is that correct?

Answer 10

@jdoe0001

Answer 11

well, keep in mind that we're just counting when the sign hops from one term to the next and changes sign so \(\bf f(x)=5x^6+4x^5+3x^4-2x^3+2x^2-x-1
\\ \quad \\ \quad \\

f(-x)=5x^6-4x^5+3x^4+2x^3+2x^2+x-1\)

Answer 12

for f( x ) I got 3
for f( -x ) another 3 too

so the real positive ones are 3 OR (3-2)
and the real negative ones are 3 OR (3-2)

thus is a 6 degree, thus 6 roots

3 positive and 3 negative
OR
3 positive and 1 negative and 2 complex
OR
1 positive and 3 negative and 2 complex

Answer 13

Thanks! :)

Answer 14

yw