Factor this polynomial into the product of two binomials (30 points): me understand how to do this. this is a challenge problem for our class, and i can't figure out how to factor it with the exponents as variables. you!

Step-by-step explanation:

Let's begin by factoring x^2 + 9x  - 10

When you do that you get

(x - 1) (x + 10)

If that can be done then let x^(2n) + 9^n - 10 be factored into

(x^n - 1) (x^n + 10) Both of these can be equated to zero.

x^n - 1 = 0

x^n = 1

Take the nth root of both sides

No matter what n is the root of 1 is still going to be 1

x^n + 10 = 0

x^n = - 10

This one is a little harder. It depends whether n is odd or even. It it is odd, no problem.  The nth root of - 10 can be found.

For example, suppose n = 5. Then the 5th root of -10 is - 1.58

You can find this on your calculator by doing

(-)

10

y^x or x^y or ^ [whichever your calculator has]

=

and you should get - 1.58

If n is even, you have a problem. If you do not know what complex numbers are, they this problem has not meaning. If n = 6, then you can take the sixth of 10. Then the answer becomes If you do know what then the answer becomes  1.468i. I would say if n is even there is no solution.

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answer: d.) the total number of protons and neutrons is 15

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