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Factor completely 2x^4y^3 - 12x^3y^2 - 8x^2y

A.) 2x^2y(x^2y^2− 6xy − 4)
B.) 2(x^4y^3 − 6x^3y^2 − 4x^2y)
C.) 2y(x^4y^2− 6x^3y− 4x^2)
D.) Prime

Respuesta :

sahkfam

Answer:

[tex]2x^2y\left(x^2y^2-6xy-4\right)[/tex]

Step-by-step explanation:

Hey!

Let's start by applying some exponent rules:

[tex]2x^4y^3-12x^3y^2-8x^2y[/tex]

[tex]=2x^2x^2yy^2-12x^2xyy-8x^2y[/tex]

Rewriting our -8 and -12

[tex]=2x^2x^2yy^2+6\cdot \:2x^2xyy+4\cdot \:2x^2y[/tex]

Factoring out the common term:

[tex]=2x^2y\left(x^2y^2-6xy-4\right)[/tex]

Thus, our final answer is [tex]\boxed{2x^2y\left(x^2y^2-6xy-4\right)}[/tex]

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