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X2 - 5/ 3x2-5x-2  +    X+1/ 3x-6

Guest Jan 11, 2018
 #1
avatar+88899 
+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)

 

 

cool cool cool

CPhill  Jan 11, 2018
edited by CPhill  Jan 11, 2018
 #2
avatar
+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
avatar+88899 
+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

 

 

cool cool cool

CPhill  Jan 11, 2018
 #4
avatar
+1

Thank you!!!smiley

Guest Jan 11, 2018

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