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Mathematics 16 Online
OpenStudy (anonymous):

someone please tell me what i am doing wrong multiply and simplify z² 8z+24 ----- *--------- 9z+27 8z^9

OpenStudy (m):

need to show your work first

OpenStudy (anonymous):

factor out a (z+3) in the numerator and denominator, leaving 8/9 cancel z^2 and leave z^7 in denominator

OpenStudy (amistre64):

simplify everything to factors to see what you can cross out

OpenStudy (anonymous):

z^2(8z+24)=z^2*8(z+3) (9z+27)(8z^9)=8(z+3)*8*z^9 8z^2(z+3) 8z^2(z+3) * 1/8z^9 1*1/8z^9 1/8z^9

OpenStudy (amistre64):

is that sposed to be an equals sign between them?

OpenStudy (anonymous):

yes this is the equation z² * 8z+24 9z+27 8z^9

OpenStudy (amistre64):

\[\frac{z z}{9 (z+3)} * \frac{8(z+3)}{8zzzzzzzzz}\]

OpenStudy (amistre64):

get rid of 2 z's top and bottom; and the (z+3) top and bottom to get; and the 8's top to bottom to get: 1/9z^6

OpenStudy (amistre64):

ack!!... 1/9z^7 :) forgot how to subtract

OpenStudy (amistre64):

does that make sense?

OpenStudy (anonymous):

i think so will u show the work so i can make notes for the next problem

OpenStudy (amistre64):

i did show the work.... cant really see how to show any more of it :/

OpenStudy (anonymous):

in my work i think i forgot about the 9

OpenStudy (amistre64):

when everything is factored; you simple cancel out (its a bad term really) everything that is the same top to bottom; because they go to 1 z/z = 1 8/8 = 1 (z+3)/(z+3) = 1 so they become redundant in the equation

OpenStudy (amistre64):

whats left is: 1 ------ 9 z^7

OpenStudy (anonymous):

ty that makes more sense

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