How do you factor #6r^3-9r^3-4r+6#?

Answer 1

Avery Alexander

I think the question contains an error: #-9r^3# should be #-9r^2#

On the assumption that I'm correct, the cubic polynomial factors easily by grouping:

#6r^3-9r^2-4r+6#

#=(6r^3-9r^2)-(4r-6)#

#=3r^2(2r-3)-2(2r-3)#

#=(3r^2-2)(2r-3)#

#=(sqrt(3)r-sqrt(2))(sqrt(3)r+sqrt(2))(2r-3)#

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Answer 2

Ethan Adkins

To factor the expression (6r^3 - 9r^3 - 4r + 6), you can group like terms and factor out common factors:

(6r^3 - 9r^3 - 4r + 6 = (6r^3 - 9r^3) + (-4r + 6))

Factor out common factors from each group:

(6r^3 - 9r^3 = 3r^3(2 - 3) = -3r^3)

(-4r + 6 = -2(2r - 3))

So the factored expression is (-3r^3 - 2(2r - 3)).

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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.