We know that an equation of degree 'n' can have 'n' solutions but does it imply that it must have 'n' solutions including complex solutions or could it have less than 'n' solutions totally as well? In the latter case is it like 1 solution repeats to become two solutions or something like that? Why should a degree 'n' equation have precisely 'n' solutions??
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It depends on how you count. For example, think for yourself about how many solutions the equation $x^n=0$ has. – Thorgott Mar 28 '22 at 17:55
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2Wikipedia on Fundamental theorem of algebra: "The theorem is also stated as follows: every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots." – VTand Mar 28 '22 at 17:58
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1A polynomial of degree $n$ may have even infinitely many roots, e. g., over the quaternions, see here. It is important to say over which domain we consider the "equation". You have tagged "complex numbers", but one should mention it also in the text. – Dietrich Burde Mar 28 '22 at 18:15
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Wouldn't counting with multiplicity make it a number less than 'n' technically? – Lumbini Ashutosh Tambat Mar 29 '22 at 11:03
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A polynomial of degree n>=1 with leading coefficient 1 is always the product of factors of the form (x + a) and (x^2 + ax + b). The square factors produce two conjugate complex solutions (that is a + bi and a - bi). Two or more of the factors can be identical so you get the same solution twice or more often. If you count them multiple times you get n solutions.
gnasher729
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Then, an n degree equation doesn't necessarily have 'n' 'distinct' roots? – Lumbini Ashutosh Tambat Mar 29 '22 at 10:59