Question

1. - / 1 points simplify the rational expression. \\(\frac{y^3 - 9y}{y^2 + 3y - 18}\\) \\(\quad,\quad y \

eq 3\\) resources read it

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \( y^3 - 9y \). We can factor out a common factor of \( y \) first:
\( y^3 - 9y = y(y^2 - 9) \)
Then, notice that \( y^2 - 9 \) is a difference of squares, which factors as \( (y - 3)(y + 3) \). So the numerator becomes:
\( y(y - 3)(y + 3) \)

Next, factor the denominator \( y^2 + 3y - 18 \). We need to find two numbers that multiply to \( -18 \) and add to \( 3 \). Those numbers are \( 6 \) and \( -3 \). So we can factor the denominator as:
\( (y + 6)(y - 3) \)

Step2: Cancel common factors

Now, the rational expression is:
\( \frac{y(y - 3)(y + 3)}{(y + 6)(y - 3)} \)
We can cancel out the common factor of \( (y - 3) \) (since \( y
eq 3 \), we don't have to worry about division by zero here):
\( \frac{y(y + 3)}{y + 6} \)
We can also expand the numerator if we want:
\( \frac{y^2 + 3y}{y + 6} \)

Answer:

\( \frac{y^2 + 3y}{y + 6} \) (or \( \frac{y(y + 3)}{y + 6} \))