16.For how many positive integers n is n3āˆ’8n2+20nāˆ’13 a prime number?



Answers:
I got the answer! There are 3 positive integers for which n^3 - 8n^2 + 20n - 13 will be a prime number. This is why...


SOLUTION.

You first have to factor n^3 - 8n2 + 20n - 13.

Well... if that was set = to 0... then what number will give you 0? That number will then be "one of the roots" of the polynomial... and, therefore, a factor of the polynomial...

Okay, try "1" and see if you get "0"... So... when n = 1...

n^3 - 8n^2 + 20n -13

= (1)^3 - 8(1)^2 + 20(1) - 13

= 1 - 8 + 20 - 13

= 0

So... you know that n = 1 is one root of the polynomial... and, therefore, the "factor" is (n - 1).

Okay... so (n - 1)(something) = the polynomial.

So... using long division.

... _n^2_- 7n_+13____
(n-1) | (n^3 - 8n^2 + 20n - 13)
... -(n^3 - n^2)
.... --------------
...... -7n^2 + 20n
...... -(-7n^2 + 7n)
.... -----------------
........ 13n - 13
........ -(13n +13)
......... ---------------
......... 0

So... when you divide n^3 - 8n^2 + 20n -13 by (n-1), you get the polynomial n^2 - 7n +13 which is the PRODUCT that results when you multiply the other two factors...

So. n^3 - 8n^2 + 20n - 13 = (n-1)(n^2 - 7n + 13)

If n (> or =) 5, then n^3 - 8n^2 + 20n -13 = (n-1)(n^2 - 7n + 13) is the product of two integers greater than 1, and thus is not prime. For n = 1, 2, 3, and 4 we have, respectively...

(1-1)(1^2 - 7*1 + 13) = (0)(7) = 0,

(2-1)(2^2 - 7*2 + 13) = (1)(3) = 3,

(3-1)(3^2 - 7*3 + 13) = (2)(1) = 2, and...

(4-1)(4^2 - 7*4 + 13) = (3)(1) = 3

Therefore, n^3 - 8n^2 + 20n - 13 is prime only when n = 2, 3, and 4.

So for 3 positive integers (n = 2, 3, and 4), n^3 - 8n^2 + 20n - 13 is a prime number.

Hope this helps!

I give you a star because this was a difficult and HARD question! Thanks for making me think! =) UGH!
lol ur homework or something?

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