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Mathematics 21 Online
OpenStudy (anonymous):

(2x²-1x) - (3x²+2) (these are numerators)

OpenStudy (anonymous):

denominator is x³ it's just the last step in simplifying a radical equation

OpenStudy (anonymous):

expression*

OpenStudy (loser66):

\[\huge\frac{(2x^2-x)-(3x^2+2)}{x^3}\]

OpenStudy (anonymous):

ok I'm awful with the equation thing, but it's 2x^2 - x over x³ minus -1x+2 over x³

OpenStudy (loser66):

ok, the same mine, friend. since they have the same denominator, you can combine like this.but what do you want me to do?

OpenStudy (jhannybean):

\[\large \cfrac{(2x^2-x)-(3x^2+2)}{x^3}\] distribute the negative to the \(\large 3x^2+2\) first. \[\large \cfrac{2x^2-x-3x^2-2}{x^3}\] Simplify the numerator. \[\large \cfrac{(2x^2-3x^2)-(x+2)}{x^2}\] What do you ge when you simplify the numerator?

OpenStudy (anonymous):

It's probably easier to explain if you just combine like terms then factor.

OpenStudy (jhannybean):

That's for the asker to do, not me.

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