How do you simplify the rational expression: #(x+2)/(4x-8)(3x-9)/(x+4)(2x-21)/(x^2-x-6)#?

1 Answer
KillerBunny
Oct 14, 2015

#(6x-63)/(4 x^2+8 x-32)#

Explanation:

The idea is to factor out numbers is the first-degree polynomial (if possible), and to factor the highest-degree, finding their roots (if possible). So, let's work separately on the three pieces:

  • First fraction:
    Numerator: #x+2-># nothing to do;
    **Denominator: #4x-8-># can factor a #4#, obtaining #4(x-2)#.

  • Second fraction:
    Numerator: #3x-9-># can factor a #3#, obtaining #3(x-3)#;
    **Denominator: #x+4-># nothing to do.

  • First fraction:
    Numerator: #2x-21-># nothing to do;
    **Denominator: #x^2-x-6-># its roots are #3# and #-2#, so we can write it as #(x-3)(x+2)#.

Writing back the whole expression with this changes gives

#color(red)(cancel(x+2))/(4(x-2)) * (3color(blue)cancel((x-3)))/(x+4) * (2x-21)/(color(blue)cancel((x-3))color(red)cancel((x+2)))#

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