poopey
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What is the product?

[tex]\[ \left(-2d^2 + s\right)\left(5d^2 - 6s\right) \][/tex]

A. \(-10d^4 + 17d^2s - 6s^2\)

B. \(-10d^4 + 17d^4s^2 - 6s^2\)

C. \(-10d^4 - 7d^2s - 6s^2\)

D. [tex]\(-10d^4 + 17d^2s + 6s^2\)[/tex]



Answer :

To find the product of the polynomials \((-2 d^2 + s)\) and \((5 d^2 - 6 s)\), 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. Let’s break this down step-by-step:

1. Distribute \(-2 d^2\) across \(5 d^2 - 6 s\):
[tex]\[ (-2 d^2) \cdot (5 d^2 - 6 s) \][/tex]
This gives us:
[tex]\[ (-2 d^2) \cdot (5 d^2) + (-2 d^2) \cdot (-6 s) \][/tex]
Calculate each term:
[tex]\[ (-2 d^2) \cdot (5 d^2) = -10 d^4 \][/tex]
[tex]\[ (-2 d^2) \cdot (-6 s) = 12 d^2 s \][/tex]

2. Distribute \(s\) across \(5 d^2 - 6 s\):
[tex]\[ s \cdot (5 d^2 - 6 s) \][/tex]
This gives us:
[tex]\[ s \cdot (5 d^2) + s \cdot (-6 s) \][/tex]
Calculate each term:
[tex]\[ s \cdot (5 d^2) = 5 d^2 s \][/tex]
[tex]\[ s \cdot (-6 s) = -6 s^2 \][/tex]

3. Combine all the terms from steps 1 and 2:
[tex]\[ -10 d^4 + 12 d^2 s + 5 d^2 s - 6 s^2 \][/tex]

4. Combine like terms (terms with the same variable and exponent):
[tex]\[ -10 d^4 + (12 d^2 s + 5 d^2 s) - 6 s^2 \][/tex]
Combine \(12 d^2 s\) and \(5 d^2 s\):
[tex]\[ -10 d^4 + 17 d^2 s - 6 s^2 \][/tex]

So, the product of the polynomials \((-2 d^2 + s)\) and \((5 d^2 - 6 s)\) is:
[tex]\[ -10 d^4 + 17 d^2 s - 6 s^2 \][/tex]

This matches the first given option:

[tex]\[ -10 d^4 + 17 d^2 s - 6 s^2 \][/tex]

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