Answer :
Sure! Let’s find the product of [tex]\((-2x - 9y^2)(-4x - 3)\)[/tex] step-by-step.
1. Distribute: To find the product of these two expressions, we'll distribute each term in the first expression by each term in the second expression. This is sometimes called using the distributive property or FOIL when dealing with two binomials.
2. Multiply each term:
- Multiply [tex]\(-2x\)[/tex] by [tex]\(-4x\)[/tex]:
[tex]\[
(-2x) \times (-4x) = 8x^2
\][/tex]
- Multiply [tex]\(-2x\)[/tex] by [tex]\(-3\)[/tex]:
[tex]\[
(-2x) \times (-3) = 6x
\][/tex]
- Multiply [tex]\(-9y^2\)[/tex] by [tex]\(-4x\)[/tex]:
[tex]\[
(-9y^2) \times (-4x) = 36xy^2
\][/tex]
- Multiply [tex]\(-9y^2\)[/tex] by [tex]\(-3\)[/tex]:
[tex]\[
(-9y^2) \times (-3) = 27y^2
\][/tex]
3. Combine the results: Now, we add all these products together:
[tex]\[
8x^2 + 6x + 36xy^2 + 27y^2
\][/tex]
So, the expanded expression and therefore the product of [tex]\((-2x - 9y^2)(-4x - 3)\)[/tex] is:
[tex]\[
8x^2 + 6x + 36xy^2 + 27y^2
\][/tex]
That matches one of the given choices, and it's the correct answer.
1. Distribute: To find the product of these two expressions, we'll distribute each term in the first expression by each term in the second expression. This is sometimes called using the distributive property or FOIL when dealing with two binomials.
2. Multiply each term:
- Multiply [tex]\(-2x\)[/tex] by [tex]\(-4x\)[/tex]:
[tex]\[
(-2x) \times (-4x) = 8x^2
\][/tex]
- Multiply [tex]\(-2x\)[/tex] by [tex]\(-3\)[/tex]:
[tex]\[
(-2x) \times (-3) = 6x
\][/tex]
- Multiply [tex]\(-9y^2\)[/tex] by [tex]\(-4x\)[/tex]:
[tex]\[
(-9y^2) \times (-4x) = 36xy^2
\][/tex]
- Multiply [tex]\(-9y^2\)[/tex] by [tex]\(-3\)[/tex]:
[tex]\[
(-9y^2) \times (-3) = 27y^2
\][/tex]
3. Combine the results: Now, we add all these products together:
[tex]\[
8x^2 + 6x + 36xy^2 + 27y^2
\][/tex]
So, the expanded expression and therefore the product of [tex]\((-2x - 9y^2)(-4x - 3)\)[/tex] is:
[tex]\[
8x^2 + 6x + 36xy^2 + 27y^2
\][/tex]
That matches one of the given choices, and it's the correct answer.
