Question

question
simplify: \\(\frac{20y^3 + 20y^2 + 30y}{-5y}\\)
answer attempt 1 out of 2

Explanation:

Step1: Factor out common term in numerator

First, factor out the greatest common factor (GCF) from the numerator \(20y^{3}+20y^{2}+30y\). The GCF of \(20y^{3}\), \(20y^{2}\), and \(30y\) is \(10y\). So we have:
\(20y^{3}+20y^{2}+30y = 10y(2y^{2}+2y + 3)\)
Now the expression becomes \(\frac{10y(2y^{2}+2y + 3)}{-5y}\)

Step2: Cancel out common factors

Next, cancel out the common factors in the numerator and the denominator. We can cancel out \(y\) (assuming \(y
eq0\)) and also divide \(10\) by \(- 5\).
\(\frac{10y(2y^{2}+2y + 3)}{-5y}=\frac{10}{-5}\times(2y^{2}+2y + 3)\)
Since \(\frac{10}{-5}=- 2\), we get:
\(-2(2y^{2}+2y + 3)\)

Step3: Distribute the -2

Now, distribute the \(-2\) across the terms inside the parentheses:
\(-2\times2y^{2}+(-2)\times2y+(-2)\times3=-4y^{2}-4y - 6\)

Answer:

\(-4y^{2}-4y - 6\)