The number of zeroes, the polynomial $x^{7} + x^{5} + x^{3} + x + 1$ can have is equal to .

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Solution

The correct option is A 7
The number of zeroes a polynomial is equal to the degree of the polynomial.

The degree of the polynomial $x^{7} + x^{5} + x^{3} + x + 1$ is 7.

Thus, $x^{7} + x^{5} + x^{3} + x + 1$ will have 7 zeros.

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x^{7} + x^{5} + x^{3} + x + 1

x^{7} + x^{5} + x^{3} + x + 1

The maximum number of zeroes the polynomial $x^{7} + x^{5} + x^{3} + x + 1$ can have is _______.

P (x) = {(\sqrt{x^{5} - 1} + x)}^{7} - {(\sqrt{x^{5} - 1} - x)}^{7}

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