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f(a)=((a^6+2a^5-3a^3-2a^4-a-1))/((a^5+a^2-a+1))

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dda ((a⁶+2a⁵−3a³−2a⁴−a−1)(a⁵+a²−a+1) )

Solution

a¹⁰+2a⁸+10a⁷+a⁶−2a⁵+18a⁴−2a³−8a²+2a−2(a⁵+a²−a+1)²

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Solution steps

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dda ((a⁶+2a⁵−3a³−2a⁴−a−1)(a⁵+a²−a+1) )

Apply the Quotient Rule: (fg )^′=f ′·g−g′·fg²

=dda (a⁶+2a⁵−3a³−2a⁴−a−1)(a⁵+a²−a+1)−dda (a⁵+a²−a+1)(a⁶+2a⁵−3a³−2a⁴−a−1)(a⁵+a²−a+1)²

dda (a⁶+2a⁵−3a³−2a⁴−a−1)=6a⁵+10a⁴−8a³−9a²−1

dda (a⁵+a²−a+1)=5a⁴+2a−1

=(6a⁵+10a⁴−8a³−9a²−1)(a⁵+a²−a+1)−(5a⁴+2a−1)(a⁶+2a⁵−3a³−2a⁴−a−1)(a⁵+a²−a+1)²

Expand (6a⁵+10a⁴−8a³−9a²−1)(a⁵+a²−a+1)−(5a⁴+2a−1)(a⁶+2a⁵−3a³−2a⁴−a−1): a¹⁰+2a⁸+10a⁷+a⁶−2a⁵+18a⁴−2a³−8a²+2a−2

=a¹⁰+2a⁸+10a⁷+a⁶−2a⁵+18a⁴−2a³−8a²+2a−2(a⁵+a²−a+1)²

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

f(a)=((a^6+2a^5-3a^3-2a^4-a-1))/((a^5+a^2-a+1))

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