How do you simplify #\frac { - 2x ^ { 3} + 4x ^ { 2} + 22x - 32} { x ^ { 3} + 2x ^ { 2} - 8x }#?

1 Answer
Andy Y.
Feb 1, 2017

#-(2(x^3-2x^2-11x+16))/(x(x+4)(x-2))#

Explanation:

Assuming you wrote the question correctly, there isn't much simplification you can do for this expression.

A negative two (#-2#), can be factored out of the numerator, giving:

Numerator: #-2(x^3-2x^2-11x+16)#

The denominator can be factored in two steps. First, factor out the #x# which is common to all the terms:

Denominator: #x(x^2+2x-8)#

Second, factor the second term:

Denominator: #x(x+4)(x-2)#

Now, simply place the new numerator over the new denominator, giving:

#-(2(x^3-2x^2-11x+16))/(x(x+4)(x-2))#.

NOTE: This is a little strange for an introductory algebra course. Students are usually given something that factors "nicely", reducing multiple terms. For example, if the problem had asked you this:

#(-2x^3+4x^2+22x-24)/(x^3+2x^2-3x)#

You would have been able to reduce that to this:

#-(2(x-4))/(x)=8/x-2#

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