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Rewrite [tex]$9x^2 + 3xy + 15x + 5y$[/tex] in factored form.

A. [tex]$(3x + 3)(5x + y)$[/tex]

B. [tex][tex]$(3x + 5)(3x + y)$[/tex][/tex]

C. [tex]$(3x + 5y)(3x + 1)$[/tex]

D. [tex]$(3x + y)(5x + 3)$[/tex]

Sagot :

GinnyAnsweridn
To rewrite the polynomial [tex]\( 9x^2 + 3xy + 15x + 5y \)[/tex] in factored form, follow these steps:

1. Identify and group terms: We begin by examining the given polynomial [tex]\( 9x^2 + 3xy + 15x + 5y \)[/tex].
2. Factor by grouping: To factor by grouping, we look for common factors in parts of the polynomial.

[tex]\( 9x^2 + 3xy + 15x + 5y \)[/tex]

Notice that we can group the terms as [tex]\((9x^2 + 3xy)\)[/tex] and [tex]\((15x + 5y)\)[/tex]:

[tex]\[ 9x^2 + 3xy + 15x + 5y = 3x(3x + y) + 5(3x + y) \][/tex]

Here, [tex]\(3x\)[/tex] is a common factor in the first group, and [tex]\(5\)[/tex] is a common factor in the second group.

3. Factor out the common binomial: Now, notice that [tex]\((3x + y)\)[/tex] is common in both groups:

[tex]\[ 3x(3x + y) + 5(3x + y) = (3x + y)(3x + 5) \][/tex]

The polynomial can now be written in factored form as [tex]\((3x + y)(3x + 5)\)[/tex].

Therefore, the correct answer is:
[tex]\[ (3x + 5)(3x + y) \][/tex]
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