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What is the product?

[tex]\[ (4y-3)\left(2y^2+3y-5\right) \][/tex]

A. [tex]\[ 8y^3+3y+15 \][/tex]

B. [tex]\[ 8y^3-23y+15 \][/tex]

C. [tex]\[ 8y^3-6y^2-17y+15 \][/tex]

D. [tex]\[ 8y^3+6y^2-29y+15 \][/tex]

Sagot :

GinnyAnswer
Certainly! Let's find the product of the two polynomials:

[tex]\[ (4y - 3) \left(2y^2 + 3y - 5\right) \][/tex]

To do this, we'll distribute each term in the first polynomial to each term in the second polynomial and then combine like terms.

1. First, distribute [tex]\(4y\)[/tex]:

[tex]\[ 4y \cdot 2y^2 = 8y^3 \][/tex]
[tex]\[ 4y \cdot 3y = 12y^2 \][/tex]
[tex]\[ 4y \cdot (-5) = -20y \][/tex]

2. Next, distribute [tex]\(-3\)[/tex]:

[tex]\[ -3 \cdot 2y^2 = -6y^2 \][/tex]
[tex]\[ -3 \cdot 3y = -9y \][/tex]
[tex]\[ -3 \cdot (-5) = 15 \][/tex]

3. Now combine all these results:

[tex]\[ 8y^3 + 12y^2 - 20y - 6y^2 - 9y + 15 \][/tex]

4. Combine like terms:

[tex]\[ 8y^3 + (12y^2 - 6y^2) + (-20y - 9y) + 15 \][/tex]
[tex]\[ 8y^3 + 6y^2 - 29y + 15 \][/tex]

So, the product is:

[tex]\[ 8y^3 + 6y^2 - 29y + 15 \][/tex]

From the given choices, the correct one is:

[tex]\[ 8 y^3 + 6 y^2 - 29 y + 15 \][/tex]
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