find the zeros of the polynomial x^2+39x+280 please it's urgent...do not spem....please..
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Q:p(X)=x³-21x²+39x-28
A:Let the zeroes be:-
a-d,a,a+d
a-d+a+a+d=12.
3a=12
a=4
a(a+d)(a-d)=28
a(a²-d²)=28
4(16-d²)=28
16-d²=7
-d²=-9
d=3
⏺️a-d=1
⏺️a = 4
⏺️a+d=7
ANS:- The ZEROES are:-1,4,7
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