Question

simplify.\\(
\frac{-3y^{2} + 9y - 6}{y^{2} - 8y + 7}
\\)

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(-3y^{2}+9y - 6\). We can factor out a \(-3\) first: \(-3(y^{2}-3y + 2)\). Then factor the quadratic inside the parentheses: \(y^{2}-3y + 2=(y - 1)(y - 2)\), so the numerator becomes \(-3(y - 1)(y - 2)\).

Next, factor the denominator \(y^{2}-8y + 7\). We need two numbers that multiply to \(7\) and add to \(-8\), which are \(-1\) and \(-7\). So, \(y^{2}-8y + 7=(y - 1)(y - 7)\).

Now the expression is \(\frac{-3(y - 1)(y - 2)}{(y - 1)(y - 7)}\).

Step2: Cancel common factors

We can cancel out the common factor \((y - 1)\) (assuming \(y
eq1\) to avoid division by zero). After canceling, we get \(\frac{-3(y - 2)}{y - 7}\), which can also be written as \(\frac{-3y + 6}{y - 7}\) or \(\frac{3(2 - y)}{y - 7}\) (but the simplified form with the factor canceled is more standard as \(\frac{-3(y - 2)}{y - 7}\) or \(\frac{3(2 - y)}{y - 7}\) or \(\frac{-3y + 6}{y - 7}\)).

Answer:

\(\frac{-3(y - 2)}{y - 7}\) (or equivalent forms like \(\frac{6 - 3y}{y - 7}\))