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Is it possible to factor #y=x^3 - 3x^2 - 4x + 12 #? If so, what are the factors?

Answer 1

Savannah Anderson

(x - 3 )(x - 2 )(x + 2 )

By factorising in pairs :

#[ x^2 ( x - 3 ) ] + [ - 4 ( x - 3 ) ]#

ie # x^2 (x - 3 ) - 4 ( x - 3 ) #

now there is a common factor of (x - 3 )

# (x - 3 )(x^2 - 4 ) #

# x^2 - 4 # is a difference of 2 squares.

and # x^2 - 4 = ( x - 2 )(x + 2 ) #

#rArr x^3 -3x^2 - 4x + 12 = (x - 3 )(x - 2 )(x + 2 ) #

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Answer 2

Ethan Adkins

Yes, it is possible to factor the expression (y = x^3 - 3x^2 - 4x + 12). The factors of the expression are ((x - 4)(x - 1)(x + 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.