Respuesta :
The quotient is [tex]\frac{3(y-5)}{2}[/tex]
Explanation:
The expression is [tex]\frac{2 y^{2}-6 y-20}{4 y+12} \div \frac{y^{2}+5 y+6}{3 y^{2}+18 y+27}[/tex]
Now, reciprocal the term [tex]\frac{y^{2}+5 y+6}{3 y^{2}+18 y+27}[/tex] and convert ÷ to ×
[tex]\frac{2 y^{2}-6 y-20}{4 y+12} \times \frac{3 y^{2}+18 y+27}{y^{2}+5 y+6}[/tex]
Taking the common terms out, we get,
[tex]\frac{2\left(y^{2}-3 y-10\right)}{4(y+3)} \times \frac{3\left(y^{2}+6 y+9\right)}{y^{2}+5 y+6}[/tex]
Factorizing each numerator and denominator, we get,
[tex]\frac{2(y-5)(y+2)}{4(y+3)} \times \frac{3(y+3)(y+3)}{(y+3)(y+2)}[/tex]
Multiplying the terms, we have,
[tex]\frac{6(y-5)(y+2)(y+3)^{2}}{4(y+3)^{2}(y+2)}[/tex]
Cancelling the common factors, we get,
[tex]\frac{3(y-5)}{2}[/tex]
Thus, the quotient is [tex]\frac{3(y-5)}{2}[/tex]
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