Question #17b55

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Answer 1 · 2017-10-13T17:59:30

Answer is #0#.

The given expression is #rarr#

#(p-a)(p-b)....(p-z)#

Note that the term getting subtracted from the term #p# is in alphabetical order.

Hence, the order is #rarr#

#a,b,c,d.....x,y,z#.

Now, in the aforementioned order, one term is itself #p#.

Hence, one term will be #(p-p)#.

#(p-p)=0#.

Hence,

#(p-a)(p-b)....(p-p)....(p-y)(p-z)#

#=(p-a)(p-b)...xx0xx...(p-y)(p-z)#

#=0#. (Answer).

Hope it Helps:)