Question

1. factor completely: $3y^2 - 12y - 288$

Explanation:

Step1: Factor out the GCF

First, find the greatest common factor (GCF) of the terms \(3y^{2}\), \(-12y\), and \(-288\). The GCF of 3, 12, and 288 is 3. So we factor out 3:
\(3y^{2}-12y - 288=3(y^{2}-4y - 96)\)

Step2: Factor the quadratic trinomial

Now we need to factor the quadratic trinomial \(y^{2}-4y - 96\). We need two numbers that multiply to \(- 96\) and add up to \(-4\). Let's find the factors of \(-96\):
The factors of \(-96\) that satisfy this are \(-12\) and \(8\) because \((-12)\times8=-96\) and \(-12 + 8=-4\).
So we can write \(y^{2}-4y - 96=(y - 12)(y+8)\)

Step3: Combine the factors

Putting it all together, the completely factored form of \(3y^{2}-12y - 288\) is:
\(3(y - 12)(y + 8)\)

Answer:

\(3(y - 12)(y + 8)\)