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Answer :
Sure! Let's find the product of the expressions [tex]\((-2x - 9y^2)\)[/tex] and [tex]\((-4x - 3)\)[/tex] by distributing each term.
### Step-by-step Solution:
1. Distribute each term in the first polynomial with each term in the second polynomial:
[tex]\((-2x)\)[/tex] times [tex]\((-4x)\)[/tex] gives:
[tex]\[ (-2) \times (-4) \times x \times x = 8x^2 \][/tex]
[tex]\((-2x)\)[/tex] times [tex]\((-3)\)[/tex] gives:
[tex]\[ (-2) \times (-3) \times x = 6x \][/tex]
[tex]\((-9y^2)\)[/tex] times [tex]\((-4x)\)[/tex] gives:
[tex]\[ (-9) \times (-4) \times y^2 \times x = 36xy^2 \][/tex]
[tex]\((-9y^2)\)[/tex] times [tex]\((-3)\)[/tex] gives:
[tex]\[ (-9) \times (-3) \times y^2 = 27y^2 \][/tex]
2. Combine all the terms obtained after distribution:
[tex]\[ 8x^2 + 6x + 36xy^2 + 27y^2 \][/tex]
Thus, the product of [tex]\((-2x - 9y^2)\)[/tex] and [tex]\((-4x - 3)\)[/tex] is:
[tex]\[ 8x^2 + 6x + 36xy^2 + 27y^2 \][/tex]
So, the correct answer is:
[tex]\[ 8x^2 + 6x + 36xy^2 + 27y^2 \][/tex]
This corresponds to the option: [tex]\( 8x^2 + 6x + 36xy^2 + 27y^2 \)[/tex] from the given choices.
### Step-by-step Solution:
1. Distribute each term in the first polynomial with each term in the second polynomial:
[tex]\((-2x)\)[/tex] times [tex]\((-4x)\)[/tex] gives:
[tex]\[ (-2) \times (-4) \times x \times x = 8x^2 \][/tex]
[tex]\((-2x)\)[/tex] times [tex]\((-3)\)[/tex] gives:
[tex]\[ (-2) \times (-3) \times x = 6x \][/tex]
[tex]\((-9y^2)\)[/tex] times [tex]\((-4x)\)[/tex] gives:
[tex]\[ (-9) \times (-4) \times y^2 \times x = 36xy^2 \][/tex]
[tex]\((-9y^2)\)[/tex] times [tex]\((-3)\)[/tex] gives:
[tex]\[ (-9) \times (-3) \times y^2 = 27y^2 \][/tex]
2. Combine all the terms obtained after distribution:
[tex]\[ 8x^2 + 6x + 36xy^2 + 27y^2 \][/tex]
Thus, the product of [tex]\((-2x - 9y^2)\)[/tex] and [tex]\((-4x - 3)\)[/tex] is:
[tex]\[ 8x^2 + 6x + 36xy^2 + 27y^2 \][/tex]
So, the correct answer is:
[tex]\[ 8x^2 + 6x + 36xy^2 + 27y^2 \][/tex]
This corresponds to the option: [tex]\( 8x^2 + 6x + 36xy^2 + 27y^2 \)[/tex] from the given choices.
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