Certainly! Let’s solve this problem step by step.
We need to multiply the two expressions:
[tex]\[
(-6y^2 + y)(3y - 2).
\][/tex]
To do this, we will use the distributive property (also known as the FOIL method for binomials) to multiply each term in the first polynomial by each term in the second polynomial. Here are the steps in detail:
1. Multiply [tex]\(-6y^2\)[/tex] by [tex]\(3y\)[/tex]:
[tex]\[
-6y^2 \cdot 3y = -18y^3.
\][/tex]
2. Multiply [tex]\(-6y^2\)[/tex] by [tex]\(-2\)[/tex]:
[tex]\[
-6y^2 \cdot -2 = 12y^2.
\][/tex]
3. Multiply [tex]\(y\)[/tex] by [tex]\(3y\)[/tex]:
[tex]\[
y \cdot 3y = 3y^2.
\][/tex]
4. Multiply [tex]\(y\)[/tex] by [tex]\(-2\)[/tex]:
[tex]\[
y \cdot -2 = -2y.
\][/tex]
Now, combine all these results together:
[tex]\[
-18y^3 + 12y^2 + 3y^2 - 2y.
\][/tex]
Next, we combine like terms:
1. Combine the [tex]\(y^2\)[/tex] terms:
[tex]\[
12y^2 + 3y^2 = 15y^2.
\][/tex]
So, the final expression is:
[tex]\[
-18y^3 + 15y^2 - 2y.
\][/tex]
This is the expanded form of the product of [tex]\((-6y^2 + y)\)[/tex] and [tex]\((3y - 2)\)[/tex].
