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Why do we get $2$ solutions for a quadratic equation and $3$ solutions for a cubic equation and $4$ for biquadratic equation and so forth?

Martin Sleziak
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Fawad
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    "At most..." . A polynomial equation of degree $;n;$ has at most $;n;$ different solutions in some given field. This has to do with the fact that $;F[x];,;;F;$ a field, is an Euclidean domain and thus UFD and etc. – DonAntonio Oct 08 '16 at 15:19
  • @DonAntonioyou are wrong, solutions are not always different. An equation can have 3 solutions but same! – Fawad Oct 08 '16 at 15:21
  • Funny, I thought you were the one who asked. Never mind: I am not wrong (and apparently you didn't read all what I wrote or you didn't understand) , but if you decide I am then enjoy. – DonAntonio Oct 08 '16 at 15:22
  • Related (although not the same): http://math.stackexchange.com/questions/25822/how-to-prove-that-a-polynomial-of-degree-n-has-at-most-n-roots and http://math.stackexchange.com/questions/219115/how-do-i-prove-that-a-polynomial-fx-of-degree-n-has-at-most-n-roots – Martin Sleziak Oct 09 '16 at 07:22

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Why do we get $2$ solutions for a quadratic equation and $3$ solutions for a cubic equation and $4$ for biquadratic equation and so forth?

This is only true if you allow the solutions to be complex. This is given by The Fundamental Theorem of Algebra, which states (quoted from the Wiki) that

"...every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ roots."

Bobson Dugnutt
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  • So if any polynomial have degree of $π$ how many roots will it have? – Fawad Oct 08 '16 at 15:28
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    @Ramanujan Then it isn't a polynomial, and the fundamental theorem doesn't hold. For it to be a polynomial, you must have non-negative integer exponents only. – Bobson Dugnutt Oct 08 '16 at 17:10
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This is because of the factor theorem: if $r$ is a root of polynomial $P(x)$, then $P(x)$ must factor as $$P(x) = (x-r)Q(x),$$ where the degree of $Q$ is one less than the degree of $P$.

Then we have $P(x) =0$ if and only if $$(x-r)Q(r) =0,$$ which occurs when $x=r$ or when $x$ is a root of $Q$. So $P$ has at most one more root than $Q$, or exactly one more if you are willing to count multiple roots multiple times.

Since a 1st-degree polynomial evidently has 1 root, an $n$th-degree polynomial has $n$ roots (or "at most $n$" if you don't count the multiple roots separately.)

MJD
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