Question

What is the quotient?

Answer 1

The quotient is $\frac{3(y-5)}{2}$

Explanation:

The expression is $\frac{2 y^{2}-6 y-20}{4 y+12} \div \frac{y^{2}+5 y+6}{3 y^{2}+18 y+27}$

Now, reciprocal the term $\frac{y^{2}+5 y+6}{3 y^{2}+18 y+27}$ and convert ÷ to ×

$\frac{2 y^{2}-6 y-20}{4 y+12} \times \frac{3 y^{2}+18 y+27}{y^{2}+5 y+6}$

Taking the common terms out, we get,

$\frac{2\left(y^{2}-3 y-10\right)}{4(y+3)} \times \frac{3\left(y^{2}+6 y+9\right)}{y^{2}+5 y+6}$

Factorizing each numerator and denominator, we get,

$\frac{2(y-5)(y+2)}{4(y+3)} \times \frac{3(y+3)(y+3)}{(y+3)(y+2)}$

Multiplying the terms, we have,

$\frac{6(y-5)(y+2)(y+3)^{2}}{4(y+3)^{2}(y+2)}$

Cancelling the common factors, we get,

$\frac{3(y-5)}{2}$

Thus, the quotient is $\frac{3(y-5)}{2}$