solve by synthetic division
(2m3-3m+10)÷(m-5)
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Answer:
Given
m−5
2m
2
−3m+10
Now (2m
2
−3m+10) can be written after dividing by (m−5) as
(2m
2
−3m+10)=((2m+7)×(m−5))+45
so 45 is remainder and (2m+7) is dividend.
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