Question

factor completely.
-5v² - 12v + 9

Explanation:

Step1: Factor out -1

First, factor out -1 from the quadratic expression to make the leading coefficient positive. So we have:
$$-1(5v^{2}+12v - 9)$$

Step2: Factor the quadratic inside the parentheses

Now, we need to factor \(5v^{2}+12v - 9\). We look for two numbers that multiply to \(5\times(-9)=-45\) and add up to 12. The numbers are 15 and -3.
Rewrite the middle term using these numbers:
$$5v^{2}+15v - 3v - 9$$

Step3: Group and factor

Group the first two terms and the last two terms:
$$(5v^{2}+15v)+(-3v - 9)$$
Factor out the greatest common factor from each group:
$$5v(v + 3)-3(v + 3)$$
Now, factor out the common binomial factor \((v + 3)\):
$$(5v - 3)(v + 3)$$

Step4: Combine with the -1 factored out earlier

Recall that we factored out -1 in Step 1, so the completely factored form is:
$$-1(5v - 3)(v + 3)=-(5v - 3)(v + 3)$$
We can also rewrite \(-(5v - 3)\) as \((-5v + 3)\) or \((3 - 5v)\), so another form is \((3 - 5v)(v + 3)\) (both are correct, but let's check the first form).
Wait, let's verify by expanding \(-(5v - 3)(v + 3)\):
First, expand \((5v - 3)(v + 3)=5v^{2}+15v-3v - 9 = 5v^{2}+12v - 9\), then multiply by -1: \(-5v^{2}-12v + 9\), which matches the original expression.

Answer:

\(-(5v - 3)(v + 3)\) (or \((3 - 5v)(v + 3)\))