Using the distributive property to find the product [tex] (y - 4x) (y^2 + 4y + 16) [/tex] results in a polynomial of the form [tex] y^3 + 4y^2 + ay - 4xy^2 - ax y - 64x [/tex]. What is the value of [tex] a [/tex] in the polynomial?

A. 4
B. 8
C. 16
D. 32



Answer :

To determine the value of [tex]\( a \)[/tex] in the polynomial resulting from the product [tex]\((y - 4x)(y^2 + 4y + 16)\)[/tex], we will use the distributive property to expand the expression step by step.

1. First, expand the expression using the distributive property:

[tex]\[ (y - 4x)(y^2 + 4y + 16) \][/tex]

This means we need to distribute each term in the first parenthesis to every term in the second parenthesis.

2. Distribute [tex]\( y \)[/tex]:

[tex]\[ y \cdot (y^2 + 4y + 16) = y \cdot y^2 + y \cdot 4y + y \cdot 16 \][/tex]

[tex]\[ = y^3 + 4y^2 + 16y \][/tex]

3. Distribute [tex]\(-4x \)[/tex]:

[tex]\[ -4x \cdot (y^2 + 4y + 16) = -4x \cdot y^2 + -4x \cdot 4y + -4x \cdot 16 \][/tex]

[tex]\[ = -4xy^2 - 16xy - 64x \][/tex]

4. Combine all the terms:

[tex]\[ (y - 4x)(y^2 + 4y + 16) = y^3 + 4y^2 + 16y - 4xy^2 - 16xy - 64x \][/tex]

5. Group the like terms together to obtain the polynomial in the desired form:

[tex]\[ y^3 + 4y^2 + 16y - 4xy^2 - 16xy - 64x \][/tex]

Rewriting this in a form that clearly groups the coefficients and like terms:

[tex]\[ y^3 + 4y^2 + ay - 4xy^2 - ax y - 64x \][/tex]

6. Comparing the polynomial obtained from the expansion with the polynomial [tex]\( y^3 + 4y^2 + ay - 4xy^2 - ax y - 64x \)[/tex], we see that the coefficients of the like terms must match. Specifically, the term involving [tex]\( y \)[/tex] and the term involving [tex]\( xy \)[/tex] need to be analyzed:

- The coefficient of [tex]\( y \)[/tex] in the expanded form is [tex]\( 16y \)[/tex].
- The corresponding term in the form we want is [tex]\( ay \)[/tex].

Hence:

[tex]\[ a = 16 \][/tex]

Therefore, the value of [tex]\( a \)[/tex] in the polynomial is [tex]\( \boxed{16} \)[/tex].