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Mathematics 22 Online
OpenStudy (anonymous):

Find the polynomial equation with real coefficients that has the given zeros.. 1-5i and 1+5i

OpenStudy (anonymous):

not sure if it's right... but I'll do it this way... [x-(1-5i)][x-(1+5i)] then just expand it out and collect like terms and use i^2=-1 = x^2-2x+24

OpenStudy (anonymous):

x^2 - 2x + 26 *

OpenStudy (anonymous):

ah.. yes yes 26.. sorry XD

OpenStudy (anonymous):

can you tell me how you got that ?

OpenStudy (anonymous):

(x-1+5i)(x-1-5i) =x²-x-5ix-x+1+5i+5ix-5i-25i² =x²-2x+1-25i² =x²-2x+1+25 =x²-2x+26 I don't know if there's a faster way...

OpenStudy (anonymous):

THANK YOU SOOOOO MUCH! I get it!!!

OpenStudy (anonymous):

Yay! Happy to help ^o^

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