How do you factor #120w ^ { 3} - 138w ^ { 2} + 36w# completely?

1 Answer
Nghi N
Jun 25, 2017

6(w - 1)(20w - 3)

Explanation:

#f(w) = 6y = 6(20w^2 - 23w + 3).#
Factor: #y = 20w^2 - 23w + 3#.
Since a + b + c = 0,
One real root is 1 and the factor is (w - 1),
One factor is #c/a = 3/20# and the factor is #(w - 3/20)#.
Factored form:
#y = 20(w - 1)(w -3/20) = (x - 1)(20w - 3)#
#f(x) = 6y = 6(w - 1)(20w - 3)#

Related questions
Impact of this question
652 views around the world
You can reuse this answer
Creative Commons License