Question

7 – 2x

For x^7-2x^6 + 3x^2 – 2x + 5 = 0, state the number of complex roots, the possible numberof real roots, and the possible rational roots @michele_laino

Answer 1

I have no idea where to start @zepdrix

Answer 2

I'm thinking...

Answer 3

why we have 7-2x?

Answer 4

Sorry thats not suppose to be there

Answer 5

if we apply the fundamental theorem of algebra, we can say that such eqauton has \(7\) roots

Answer 6

Right i know that part but how would i do the rest

Answer 7

@Michele_Laino

Answer 8

if we replace \(x=-2\), we get:
\[p\left( { - 2} \right) = - {2^7} - 2 \cdot {\left( { - 2} \right)^6} + 3 \cdot {2^2} - 2 \cdot \left( { - 2} \right) + 5 = - 235 < 0\]
whereas, if we replace \(x=-1\), we get:
\[p\left( { - 1} \right) = - {1^7} - 2 \cdot {\left( { - 1} \right)^6} + 3 \cdot {1^2} - 2 \cdot \left( { - 1} \right) + 5 = 7 > 0\]
so, we can say, that we have one real root, inside this interval \((-2,-1)\), since any polynomial function, is a continuous fnction

Answer 9

5 complex roots; 1, 3, or 5 real roots

possible rational roots: ±1, ±5

7 complex roots; 1, 3, 5, or 7 real roots

possible rational roots: ±1, ±5

7 complex roots; 2, 4, or 6 real roots

possible rational roots: ±1, ±5

5 complex roots; 1, 3, or 5 real roots

possible rational roots: ±, ±1, ±5
These are my answer choices @Michele_Laino

Answer 10

@Michele_Laino

Answer 11

what do you think it would be

Answer 12

we can write this:
\[p\left( 1 \right) \ne 0,\quad p\left( 3 \right) \ne 0,\quad p\left( 5 \right) \ne 0\]

Answer 13

I think that we have \(6\) complex roots and only one real root

Answer 14

Ok so what would the answer be

Answer 15

@Michele_Laino

Answer 16

sincerely, I don't know, since I don't know if the real root can be written as a rational root or not as a rational root, furthermore, there is not the option with 6 complex roots

Answer 17

I know that is why im confused

Answer 18

please look at this graph: