SOLUTION: factorise: 6(2x-3)^2-5(2x-3)-4

Question 1113069: factorise:
6(2x-3)^2-5(2x-3)-4
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.

1.  Introduce new variable  u = 2x-3.

    Then your polynomial takes the form  6u^2 - 5u - 4.



2.  Then you can factorize the last polynomial

    6u^2 - 5u - 4 = (3u-4)*(2u+1).                      (1)



3.  Now substitute back (2x-3) instead of "u" into (1).  You will get

    6(2x-3)^2-5(2x-3)-4 = (3*(2x-3)-4)*(2*(2x-3)+1)     (2)



4.  Your next and final step is to simplify (2):

     6(2x-3)^2-5(2x-3)-4 = (3*(2x-3)-4)*(2*(2x-3)+1) = (6x-13)*(4x-5).



Answer.  6(2x-3)^2-5(2x-3)-4 = (6x-13)*(4x-5).

Solved.

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