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OpenStudy (anonymous):

one last question and im going to bed can factorization help solve a quadratic equation if it is made up of a prime polynomial

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OpenStudy (anonymous):

yes it can help if u try to do so

OpenStudy (anonymous):

can you give an example

OpenStudy (anonymous):

ok wait

OpenStudy (anonymous):

no no sorry i didn't read u wrote prime polynomial AS IF THERE ARE PRIME POLYNOMIAL NOTHING CAN BE TAKEN COMMON SO ANSWER IS NO :)

OpenStudy (anonymous):

Do you mean prime coefficients and constant?

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OpenStudy (anonymous):

normal definition of "prime" means that it cannot be factored. of course you need refrence to what "ring; you are factoring over. example \[x^2-2\] is "prime" over the integers but \[x^2-2=(x+\sqrt{2})(x-\sqrt{2})\] if you are factoring over reals. similarly \[x^2+1\] is prime over reals but is \[(x+i)(x-i)\] if you allow complex numbers

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