To solve the equation [tex]\(3y^2 - 36y + 81 = 0\)[/tex] by factoring, let's first simplify the equation and look for a factorable form.
Step 1: Factor out the common term
Notice that each term in the equation has a common factor of 3. So, factor out the 3 from the equation:
[tex]\[
3(y^2 - 12y + 27) = 0
\][/tex]
Step 2: Simplify the equation
Now we have:
[tex]\[
y^2 - 12y + 27 = 0
\][/tex]
Step 3: Factor the quadratic expression
We need to factor the quadratic [tex]\(y^2 - 12y + 27\)[/tex]. We look for two numbers that multiply to [tex]\(27\)[/tex] (the constant term) and add up to [tex]\(-12\)[/tex] (the coefficient of the linear term [tex]\(y\)[/tex]).
The correct pair of numbers is [tex]\(-3\)[/tex] and [tex]\(-9\)[/tex], because:
[tex]\[
-3 \cdot -9 = 27 \quad \text{and} \quad -3 + (-9) = -12
\][/tex]
Step 4: Write the factored form
Using the pair [tex]\(-3\)[/tex] and [tex]\(-9\)[/tex], we can factor the quadratic expression as follows:
[tex]\[
y^2 - 12y + 27 = (y - 3)(y - 9)
\][/tex]
Step 5: Solve for [tex]\(y\)[/tex]
Now that we have factored the equation, we set each factor equal to zero to find the solutions for [tex]\(y\)[/tex]:
[tex]\[
(y - 3)(y - 9) = 0
\][/tex]
Setting each factor equal to zero gives us two equations:
[tex]\[
y - 3 = 0 \quad \text{or} \quad y - 9 = 0
\][/tex]
Solving these equations:
[tex]\[
y = 3 \quad \text{or} \quad y = 9
\][/tex]
Step 6: State the solution
The solutions to the equation are [tex]\(y = 3\)[/tex] and [tex]\(y = 9\)[/tex]. Therefore, the correct choice is:
A. The solution is [tex]\( y = 3, 9 \)[/tex].
