SOLUTION: Factor. Began by factoring out the common monomial 12x^2 - 68x + 80 NOTE: ^2 means x squared

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor. Began by factoring out the common monomial 12x^2 - 68x + 80 NOTE: ^2 means x squared      Log On


   

Question 1023110: Factor. Began by factoring out the common monomial
12x^2 - 68x + 80

NOTE:
^2 means x squared
Found 2 solutions by ikleyn, god2012:
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
Factor. Began by factoring out the common monomial
12x^2 - 68x + 80
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

12x%5E2+-+68x+%2B+80 = 4*(x-4)*(3x-5).


Answer by god2012(113) About Me  (Show Source):
You can put this solution on YOUR website!
12x^2 - 68x + 80
Here the factors are derived as below:
1. Try to find 2 numbers whose sum is equivalent to the middle number 68 and product is the product of the first and the last constants , in this case they are 12 and 80.
Sum = -68 = (-48) + (-20)
Product = 12*80 = 960 = 48*20 = (-48) * (-20)
12x^2 - 68x + 80 = 12x^2 - 48x - 20x + 80
= 12x(x - 4) - 20(x - 4)
= (x - 4)(12x - 20)
= (x - 4)4(3x - 5)
12x^2 - 68x + 80 = 4(x - 4)*(3x - 5)