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Question
Find whether the first polynomial is a factor of the second.
2a − 3, 10a2 − 9a − 5
Solution
\[\frac{{10a}^2 -9a-5}{2a-3}\]
\[ = \frac{5a(2a-3)+3(2a-3)+4}{2a-3}\]
\[ = \frac{(2a-3)(5a+3)+4}{2a-3}\]
\[ =(5a+3)+ \frac{4}{2a-3}\]
\[ \because \text{Remainder = 4}\]
\[ \therefore \text{( 2a-3) is not a factor of}\ {10a}^2 -9a-5.\]
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