anonymous
  • anonymous
Bet you cant solve this.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
For how many integers n with |n| <500 can the polynomial \[Pn(x)=x^{6}+n\] be written as a product of two non-constant polynomials with integer coefficients?
anonymous
  • anonymous
your right
JoannaBlackwelder
  • JoannaBlackwelder
https://www.wyzant.com/resources/answers/59611/for_how_many_integers_n_with_ini_500_can_the_polynomial_pn_x_x6_n_be_written_as_a_product_of_two_non_constant_polynomials_with_integer_coefficients

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anonymous
  • anonymous
STOP CHEATING LOL @JoannaBlackwelder
anonymous
  • anonymous
its 0 ofc.
anonymous
  • anonymous
would attack this problem along these lines which you can refine. If the polynomial Pn(x) = (x6 + n) could be written as a product of 2 non-constant polynomials with integer coefficients, then by the Fundamental Theorem of Algebra one of those factors would represent a real root when Pn(x) = 0. Let's plot Pn(x). First x6 is basically a parabola going thru (x,y) = (0,0) that opens upward and is symmetric about the y-axis. The + n is a positive or negative integer that raises or lowers the parabola along the y-axis leaving x alone. Hence, for n>=+1, Pn(x) cannot have roots since it always has positive y, and so cannot be expresses in the product format. For n=0, you get the degenerate case of x6 = 0 with 6 identical x=0 roots. so you can represent x6 = x1x5, x2x4 etc. For n negative integers, the parabola gets lowered so there are exactly 2 distinct real integer roots: If n = -1, the roots are x = +- 1 and a factorization is possible If n = -64, the roots are x = +- 2 and a factorization is possible If x = +-3, then n would be -729 = - (3)6, which exceeds your |n| < 500 constraint.

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