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# Adding and subtracting Rational expressions

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X2 - 5/ 3x2-5x-2  +    X+1/ 3x-6

Guest Jan 11, 2018
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#1
+80978
+2

[ x^2  - 5]  / [ 3x^2 - 5x - 2 ]  +   [ x + 1]   /  [ 3x  - 6]

Factor the denominators

[ x^2  - 5 ] / (  [ 3x + 1 ] [ x - 2]  )    +  [ x + 1 ] / [ 3 ( x - 2) ]

So....getting a common denominator, we have

[ 3 (x^2 - 5)   + [ x + 1] [ 3x + 1] ]  /   [  3 (x - 2) (3x + 1) ]

[ 3x^2 - 15  + 3x^2 + 3x + x + 1  ]  /  [ 3 (x - 2) (3x + 1 ) ]

[ 6x^2  + 4x - 14 ]  / [ 3 (x - 2) (3x + 1 ) ]

[ 2 ( 3x^2 + 2x - 7 ) ]  / [ 3 ( x - 2) (3x + 1) ]  =

(6 x^2 + 4 x - 14) / (9 x^2 - 15 x - 6)

CPhill  Jan 11, 2018
edited by CPhill  Jan 11, 2018
#2
+1

Thank you... how did you know it would be (3x + 1)(x-2) I am having a lot of trouble factoring

Guest Jan 11, 2018
#3
+80978
+2

3x^2  - 5x  - 2

Yeah....this can be difficult whenever the lead coefficient isn't a "1"

Let me show you a trick...it doesn't always work, but it sometines does

Write -5x  as   -6x  + 1x   and we have

3x^2 - 6x  + 1x  - 2      factor by grouping

3x ( x - 2)  + 1 ( x - 2)      the common factor is  (x - 2)....so we have

( x - 2) (3x + 1)    !!!

Here's another resource that might help..

http://www.coolmath.com/algebra/04-factoring/05-trinomials-undoing-FOIL-2-01

CPhill  Jan 11, 2018
#4
+1

Thank you!!!

Guest Jan 11, 2018

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