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Divide: $6x^{5} +4x^{4} -27x^{3} -7x^{2} -27x-6$ by $2x^{2} -3$
Given: $6x^{5} +4x^{4} -27x^{3} -7x^{2} -27x-6$ and $2x^{2} -3$
To calculate: We have to find the value of $6x^{5} +4x^{4} -27x^{3} -7x^{2} -27x-6$ by $2x^{2} -3$
Solution:
$2x^{2} \ -\ 3\ \sqrt{6x^{5} \ +\ 4x^{4} \ -\ 27x^{3} -7x^{2} -27x -6} \ \ \ \ ( \ 3x^{3} \ +\ 2x^{2} \ -\ 9\ -\frac{1}{2}$
$6x^{5} \ \ \ \ \ \ \ \ \ \ \ -9x^{3}$
$4x^{4} \ -\ 18x^{3} -7x^{2} -27x -6$
$4x^{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -6x^{2}$
$-18x^{3} \ -\ x^{2} \ -\ 27x\ -\ 6$
$-18x^{3} \ \ \ \ \ \ \ \ \ \ \ \ \ +27x\ $
$-x^{2} \ -\ 54x\ -\ 6\ \ $
$-x^{2} \ \ \ \ \ \ \ \ \ \ \ \ +\frac{3}{2} \ \ \ \ \ \ $
$-54x-\frac{15}{2} \ \ \ $
So, $\frac{6x^{5} \ +\ 4x^{4} \ -\ 27x^{3} \ -\ 7x^{2} \ -\ 27x\ -\ 6}{2x^{2} \ -\ 3} \ =\ 3x^{3} \ +\ 2x^{2} \ -\ 9x\ -\ \frac{1}{2} \ +\ \frac{-54x\ -\ \frac{15}{2}}{2x^{2} \ -\ 3}$.
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