QUESTION IMAGE
Question
what is the quotient?
\\(\frac{2y^2 - 6y - 20}{4y + 12} div \frac{y^2 + 5y + 6}{3y^2 + 18y + 27}\\)
\\(circ \frac{2}{3(y - 5)}\\)
\\(circ \frac{3(y - 5)}{2}\\)
\\(circ \frac{(y - 5)(y + 2)^2}{8(y + 3)^2}\\)
\\(circ \frac{3(y + 5)(y - 2)}{2(y + 2)}\\)
Step1: Factor each expression
First, factor the numerators and denominators:
- For the first fraction: \(2y^2 - 6y - 20 = 2(y^2 - 3y - 10) = 2(y - 5)(y + 2)\) (factoring the quadratic), and \(4y + 12 = 4(y + 3)\) (factoring out 4).
- For the second fraction: \(y^2 + 5y + 6 = (y + 2)(y + 3)\) (factoring the quadratic), and \(3y^2 + 18y + 27 = 3(y^2 + 6y + 9) = 3(y + 3)^2\) (factoring the quadratic as a perfect square).
Step2: Rewrite the division as multiplication
Dividing by a fraction is the same as multiplying by its reciprocal. So the expression becomes:
Step3: Cancel out common factors
Now, cancel out the common factors in the numerators and denominators:
- Cancel \((y + 2)\) from the first numerator and the second denominator.
- Cancel \((y + 3)\) from the first denominator and one \((y + 3)\) from the second numerator (leaving one \((y + 3)\) in the numerator, but wait, actually, let's check again:
Wait, the second numerator has \(3(y + 3)^2\), so when we multiply, the first denominator is \(4(y + 3)\), so we can cancel one \((y + 3)\) from numerator and denominator. Also, the first numerator has \(2\) and the first denominator has \(4\), which simplifies to \(\frac{2}{4}=\frac{1}{2}\), but wait, let's do it step by step:
After canceling \((y + 2)\) and one \((y + 3)\):
Wait, no, let's re - express the multiplication:
Simplify the coefficients and the remaining factors:
The \(2\) and \(4\) simplify to \(\frac{2}{4}=\frac{1}{2}\), then we have:
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\(\boldsymbol{\frac{3(y - 5)}{2}}\) (which corresponds to the option \(\frac{3(y - 5)}{2}\))