Respuesta :

gmany
[tex]\dfrac{12x^3-4x^2+8x}{-2x}=\dfrac{12x^3}{-2x}+\dfrac{-4x^2}{-2x}+\dfrac{8x}{-2x}\\\\=-6x^2+2x-4[/tex]

Answer:

-6x² + 2x - 4

Step-by-step explanation:

[tex]\frac{12x^3 - 4x^2 + 8x}{-2x}[/tex]

Step 1:

First, group the integers together and divide those:

[tex]\frac{12}{-2}[/tex] = -6

[tex]\frac{-4}{-2}[/tex] = 2 (since 4 is also a negative number, you will turn the (-) sign into a (+) sign given the rule that two negatives = a positive.)

[tex]\frac{8}{-2}[/tex] = -4

Step 2:

Next, group the variables and divide those (when in division, you will subtract the exponents instead of dividing):

[tex]\frac{x^3}{x}[/tex] = x²

[tex]\frac{x^2}{x}[/tex] = x (when there is only a variable and no exponent, you can assume 1 as it's exponent.)

[tex]\frac{x}{x}[/tex] = [tex]x^0[/tex] = 0

Step 3:

Combine the integers and variables in the order given and you will get:

-6x² + 2x - 4