Finding the positive, negative complex zeros

Question

The equation: f(x)=-13x^10-11x^8-7x^6-7

My question is I found and I believe that it is correct that there are 0 negative and/or positive roots, as I see from graphing, but I cannot tell how many complex zeros there are supposed to be. I know it is either 8 or 10 but I am not 100% sure on how to tell which it is.

Thanks!

Answer 1

p = {+-1,+-7} are the factors of the tail constant

q = {+-1, +-13} are the factors of the leading coefficient

p/q = { +-1, +-1/13, +-7, +-13} is the set of POSSIBLE rational solutions.

However, NONE of these make the polynomial zero when plugged,

So there are no rational solutions...

As stated by the other tutors, DesCartes Rule of signs verifies this

result as there are no sign changes for P(x) , nor for P(-x)

As you have stated, the graph is a flattened

parabola, belly is up and below the x-axis, going

downwards, so it does not cross the x-axis; This polynomial

function is ALWAYS negative. As a result, there are no irrational solutions either

Therefore there are no REAL solutions..

There must be 10 solutions, as it is degree 10,

and all complex/imaginary solutions MUST appear

in conjugate pairs...

so the solutions are of the form:

x = a+bi and a-bi

x = c+di and c-di

x = e+fi and e-fi

x = g+hi and g-hi

x = k+ni and k-ni

The first pair implies that there exists

a quadratic trinomial with solutions x = a+bi and x=a-bi

that divides this polynomial

the factors are (x-a) + bi and (x-a) - bi

The quadratic is then (x-a)^2 + b^2 = x^2 -2ax + a^2 + b^2

Using the same reasoning, the following four

quadratics must also divide the given polynomial:

(x-c)^2 + d^2 = x^2 -2cx + c^2 + d^2

(x-e)^2 + f^2 = x^2 -2ex + e^2 + f^2

(x-g)^2 + h^2 = x^2 -2gx+g^2 + h^2

(x-k)^2 + n^2 = x^2 - 2kx+k^2 + n^2

Multiplying these all together will give a 10th degree polynomial,

which can be found using the Undetermined Coefficients method.

there are 10 unknown coefficients a,b,c,d,e,f,g,h,k,n which shall

appear is this polynomial.

They get equated to -13, 0, -11, 0 , -7, 0, 0, 0 ,0 ,0 ,-7

So there will be at most 11 equations involving these 10 unknown

coefficients which hopefully has a solution.

Using the calculator at mathportal dot com, the 10 complex solutions are:

x1=0.53143+0.86533i

x2=0.53143−0.86533i

x3=−0.53143+0.86533i

x4=−0.53143−0.86533i

x5=−0.78995+0.32743i

x6=−0.78995−0.32743i

x7=0.78995+0.32743i

x8=0.78995−0.32743i

x9=0.97315i

x10=−0.97315i

Answer 2

By using Descartes' Rule of Signs, we will look at how many roots does the polynomial have.

f(x)= -13x¹⁰ - 11x⁸ - 7x⁶ - 7

Signs: -, -, -, - No changes in sign. Therefore, there are no positive roots.

f(-x) = -13(-x)¹⁰ - 11(-x)⁸ - 7(-x)⁶ - 7 = -13x¹⁰ - 11x⁸ - 7x⁶ - 7

Signs: -, -, -, - No changes in sign. Therefore, there are no negative roots.

For complex roots, there can be 0, 2, 4, 6, 8, or 10. Keep in mind that complex roots must come in pairs (i.e., a + bi and a - bi).

This polynomial will have 10 complex roots since the degree is 10.

Answer 3

You are correct in saying no positive or negative real numbers are zeros.

By the Fundamental Theorem of the Algebra, a polynomial of degree n has n number of zeros.

This polynomial is of degree 10, therefore 10 zeros. If none are real then 10 are complex.

Answer 4

You have to use DDecart's rule of sighns: we have 0 changes of sighn in f(x); so no positive zeros.

Because f(-x) = f(x) 0 changes of sighn in f(- x) no negative zeros. Now f(0) = - 7 ≠ 0.

So, this equation has only complex number solutions, 5 pairs of complex conjugate solutions.