9b3-18b2+8b-1/3b-2 Final an outcome : 27b3 - 54b2 + 23b - 6 ————————————————————— 3 step by step solution : action 1 : 1 leveling — 3 Equation at the finish of action 1 : 1 ...

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Is this an agree factorization that \frac23b^5-\frac16b^3+\frac49b^2-1? Why or why not?
https://www.quora.com/Is-this-an-acceptable-factorization-of-frac-2-3-b-5-frac-1-6-b-3+-frac-4-9-b-2-1-Why-or-why-not
No. Because that one thing, it’s not even correct. At b = \frac12, the initial polynomial i do not care \frac23 \frac132 - \frac16 \frac18 + \frac49 \frac14 - 1 = -\frac89, ...
\displaystyle33b^2-18 Explanation: \displaystyle\textone means of separating is to use the divisor as a element in\displaystyle\textthe numerator\displaystyle\textconsider the numerator ...
6b3-24b2/b3+b2-20b Final result : (6b3 + 13b2 + 6b + 12) • (b - 2) ———————————————————————————————— b step by step solution : action 1 : b2 simplify —— b3 splitting exponential expressions : 1.1 ...
a2-b2-4a-4b/3a-3b-12 Final result : 3a2 - 4ab - 12a - 3b2 - 9b - 36 ——————————————————————————————— 3 action by action solution : action 1 : b leveling — 3 Equation in ~ the end of action 1 : b ...

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b2-3b-10/(b-2)2b-2/b-5 Final an outcome : b5 - 7b4 + 11b3 - 4b2 - 12b - 8 ——————————————————————————————— b • (b - 2)2 Reformatting the input : changes made to her input need to not influence the solution: ...
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\left< \beginarray l l 2 & 3 \\ 5 & 4 \endarray \right> \left< \beginarray l l l 2 & 0 & 3 \\ -1 & 1 & 5 \endarray \right>

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