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Based on the Fundamental Theorem of Algebra, how many complex roots does each of the following equations have? Write your answer as a number in the space provided. For example, if there are twelve complex roots, type 12. x(x2 - 4)(x2 + 16) = 0 has a0 complex roots (x 2 + 4)(x + 5)2 = 0 has a1 complex roots x6 - 4x5 - 24x2 + 10x - 3 = 0 has a2 complex roots x7 + 128 = 0 has a3 complex roots (x3 + 9)(x2 - 4) = 0 has a4 complex roots

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nobillionaireNobley
According to the fundamental theorem of algebra, every polynomial of degree n has n complex zeroes.

Part A:

Given

[tex]x(x^2-4)(x^2+16)=0 \\ \\ \Rightarrow x(x^4+12x^2-64)=0 \\ \\ \Rightarrow x^5+12x^3-64x=0[/tex]

Thus, the given polynomial is of degree 5 and hence has 5 omplrex roots.



Part B:

Given

[tex](x^2+4)(x+5)^2 = 0 \\ \\ \Rightarrow(x^2+4)(x^2+10x+25)=0 \\ \\ \Rightarrow x^4+10x^3+25x^2+4x^2+40x+100=0 \\ \\ \Rightarrow x^4+10x^3+29x^2+40x+100=0[/tex]

Thus, the polynomial is of degree 4 and hence has 4 complex roots.



Part C:

Given

[tex]x^6-4x^5-24x^2+10x-3=0[/tex]

Thus, the given polynomial is of degree 6 and hence has 6 complex roots.



Part D:

Given

[tex]x^7+128=0[/tex]

Thus, the given polynomial is of degree 7 and hence has 7 complex roots.



Part E:

Given

[tex](x^3+9)(x^2-4)=0 \\ \\ \Rightarrow x^5-4x^3+9x^2-36=0[/tex]

Thus, the given polynomial is of degree 5 and hence has 5 complex roots.

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