Solving Z^7+128: Finding Factors & Easier Ways

  • Thread starter morbello
  • Start date
In summary, the conversation revolves around finding the seventh root of -128 and working out the factors of z^7+128. The suggested method is to use synthetic division and factoring, while also considering complex numbers and De Moivre's Theorem. The context of the problem and the use of complex numbers are also mentioned.
  • #1
morbello
73
0
ive been working on the seventh root off -128 still have not got it.

but now I am trying to work out the factors off z^7+128 do i have to work off z-2 and the quadratic z^2-z-c to get the answers or do you think there is an easyer way.



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The Attempt at a Solution

 
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  • #2


I believe the seventh root of -128 is -2, so the factor would be (x + 2), so use synthetic division to see what you have left, then work from there.
 
  • #3


Hrm. I would have suggested factoring 128 rather than just giving him that answer.

For the O.P. could you give more context? Is there a specific form for the answers you're looking for (e.g. product of linears and quadratics that don't have real roots)? Do you know about complex numbers?
 
  • #4


im learning so all help is ok.thank you for your help.
 
  • #5


There should be only one real root to z^7 + 128. This occurs at a z of -2. The other 6 roots are complex roots that can be determined by De Moivre's Theorem, the first of which is approximately: 1.80 + 0.87i.
 

FAQ: Solving Z^7+128: Finding Factors & Easier Ways

How do I solve Z^7+128?

In order to solve Z^7+128, you need to factor the polynomial. This means finding the numbers or expressions that can be multiplied together to get Z^7+128. Once you have factored the polynomial, you can easily find the solutions.

Can I use the quadratic formula to solve Z^7+128?

No, the quadratic formula is only applicable to polynomials of degree 2 (exponents up to 2). Z^7+128 is a polynomial of degree 7, so the quadratic formula cannot be used to solve it. Instead, you will need to use other methods such as factoring or synthetic division.

Is there an easier way to find the factors of Z^7+128?

Yes, there are a few techniques that can make factoring Z^7+128 easier. One method is to look for common factors between the terms of the polynomial. Another method is to use the rational roots theorem to narrow down the possible factors. Additionally, you can use the grouping method to factor the polynomial into smaller, more manageable parts.

What are the solutions to Z^7+128?

The solutions to Z^7+128 are the values of Z that make the polynomial equal to zero. These values can be found by factoring the polynomial and setting each factor equal to zero. The resulting values will be the solutions to the polynomial.

Can I use a calculator to solve Z^7+128?

While a calculator can be helpful in computing the values of Z^7+128, it cannot solve the polynomial for you. You will still need to use factoring or other techniques to find the solutions. Using a calculator can make the process quicker and easier, but it is not a substitute for understanding the concepts and methods used to solve the polynomial.

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