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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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