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How do you multiply #(3x^2-1)/(4x(x-5)) div ((3x^2)(x-1))/(x-5)#?

Answer 1

First, remember that when you're dividing two fractions, you invert the other and multiply them instead, like this:

#(3x^2-1)/(4x(x-5))*(x-5)/((3x^2)(x-1))#

We can already cancel some things that are both dividing and multiplying the equation and, thus, having a null effect in it.

#(3x^2-1)/(4xcancel(x-5))*cancel(x-5)/((3x^2)(x-1))#

#(3x^2-1)/(12x^4-12x^3)#

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Answer 2

Eva Adamson

To multiply the given expressions, you need to multiply the numerators and denominators separately.

First, multiply the numerators: (3x^2-1) * (3x^2)(x-1) = 9x^4 - 3x^2 - 3x^2 + 1 = 9x^4 - 6x^2 + 1.

Next, multiply the denominators: (4x(x-5)) * ((3x^2)(x-1))/(x-5) = 12x^3(x-1).

Finally, simplify the expression by dividing the numerator by the denominator: (9x^4 - 6x^2 + 1) / 12x^3(x-1).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.