question_answer

Divide : \[15\left( y+3 \right)\left( {{y}^{2}}-16 \right)\text{ }by\text{ }5\text{ }\left( {{y}^{2}}-y-12 \right)\]

Answer:

Factorising \[15\left( y+3 \right)\text{ }\left( {{y}^{2}}-16 \right),\]

We get, \[5\times 3\times \left( y+3 \right)\left( y-4 \right)\left( y+4 \right)\]

On factorising, \[5\left( {{y}^{2}}-\text{ }y-12 \right),\]we get

\[5\left( {{y}^{2}}-4y+3y-12 \right)\]

\[=5\left[ y\left( y-4 \right)+3\left( y-4 \right) \right]\]

\[=5\left( y-4 \right)\left( y+3 \right)\]

Therefore, on dividing the first expression by the second expression, we get \[\frac{15(y+3)({{y}^{2}}-16)}{5({{y}^{2}}-y-12)}\]

\[=\frac{5\times 3\times (y+3)(y-4)(y+4)}{5\times (y-4)(y+3)}\]

\[=3\left( y+4 \right)~~\]