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Mathematics

1 answer:

My name is Ann [436]4 years ago

4 0

It is the correct answer
Hope this help u

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

Part A:

Given

$x(x^2-4)(x^2+16)=0 \\ \\ \Rightarrow x(x^4+12x^2-64)=0 \\ \\ \Rightarrow x^5+12x^3-64x=0$

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Part B:

Given

$(x^2+4)(x+5)^2 = 0 \\ \\ \Rightarrow(x^2+4)(x^2+10x+25)=0 \\ \\ \Rightarrow x^4+10x^3+25x^2+4x^2+40x+100=0 \\ \\ \Rightarrow x^4+10x^3+29x^2+40x+100=0$

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

Part C:

Given

$x^6-4x^5-24x^2+10x-3=0$

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

Part D:

Given

$x^7+128=0$

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

Part E:

Given

$(x^3+9)(x^2-4)=0 \\ \\ \Rightarrow x^5-4x^3+9x^2-36=0$

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

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3 years ago