Question

How many complex roots does the function have? Then list the possible rational roots.

Question: 2x^4-x^3=2x^2+5x-26= 0

Please help me by showing step by step. I would like to learn how to get the answer!!

Answer 1

you mean -2x^2?

Answer 2

=2x^2

Answer 3

sorry!!!! it was meant to be a plus sign

2x^4-x^3+2x^2+5x-26= 0

Answer 4

oh, ok.. so you look at the constant, 26, try to divide the polynomial by factors of 26, like (x+1), (x-1), (x+2), (x-2), (x+13), (x-13)..

Answer 5

if the division has no remainder(zero) it's a factor for this polynomial

Answer 6

I tried all the factors and they didn't work

Answer 7

yeah, I'm looking at the graph, it has no 'clean' roots

Answer 8

so what would be the best way to find this? i saw that i could use the quadratic equation but this question has 5 terms... i thought you can only use the quadratic equation if you had 3 terms (ax^2+bx+c)

Answer 9

maybe you could factor something and get quadratic terms..

Answer 10

I don't know how to solve this, if I figure it out I will tell you

Answer 11

thanks for trying

Answer 12

Are you 100% sure you have the right equation?

Answer 13

As it is, it is irreducible (cannot be factored).

Answer 14

@whpalmer4 do you know how to find non-integer roots?

Answer 15

Yeah, I type N[Solve[2x^4-x^3+2 x^2+5x-26==0,x]] and Mathematica spits them back at me :-)

Answer 16

My interpretation of this problem is that we're supposed to use Descartes' Rule of Signs to figure out the possible combinations of root types (real positive, real negative, complex). Then use the rational root theorem to list the possible rational roots, which you've sort of already done (but it didn't look like you accounted for the fact that the leading term had a coefficient other than 1).

Answer 17

ok..

Answer 18

So our polynomial is \[f(x) = 2x^4-x^3+2x^2+5x-26\]
To apply DRoS, we write out the polynomial in descending order of exponents, as we already have it:\[2x^4-x^3+2x^2+5x-26\]Now scanning from left to right, we count the number of sign changes:
2 to -1 is 1 sign change
-1 to 2 is 1 sign change
2 to 5 is 0 sign change
5 to -26 is 1 sign change
---------------------
3 sign changes, so there are either 3 or 3-2n positive real roots

Next, do the same with \(f(-x)\) which is easily constructed by writing \(f(x)\) and changing the signs on the terms with odd exponents:
\[f(-x) = 2x^4+x^3+2x^2-5x-26\]
Count the sign changes again:
2 to 1 is 0 sign change
1 to 2 is 0 sign change
2 to -5 is 1 sign change
-5 to -26 is 0 sign change
---------------------
1 sign change, so there is 1 negative real root

We know that in a polynomial with only real coefficients, any complex roots must come in complex conjugate pairs, \(a \pm bi\)

We also know that in a single-variable polynomial of degree \(n\) we must have \(n\) roots in total. Our possibilities here are

3 positive real roots, 1 negative real root, 5-(3+1) = 1 complex root
but that isn't allowable because the complex roots must come in pairs
3-2n = 3-2(1) = 1 positive real root, 1 negative real root, 2 complex roots = 4 roots
that's the only possible configuration of roots for this polynomial

Answer 19

with a higher degree, we could end up with a situation where the positive or negative real root count could be a number of different values, not just 2 or 1 like we had here. In those cases, you may not be able to definitively establish the root structure without finding them...

Answer 20

Here are the actual solutions in symbolic and numeric form. Also a graph of the region where the real roots appears.

Answer 21

Aren't you glad you don't actually have to find those suckers? :-)

Answer 22

x_X this is calculus?