How do you factor #6x^3-x^2+17=2x^2+47#?

1 Answer
Apr 20, 2018

Answer:

#x^2 (2x - 1) = 10#

Explanation:

Factor
#6x^3−x^2+17=2x^2+47#

1) Subtract #2x^2# from both sides to get all the #x# terms together
#6x^3 - x^2 - 2x^2 + 17 = 47#

2) Subtract #17# from both sides to get all the constants on the same side
#6x^3 - x^2 - 2x^2 = 47 - 17#

3) Combine like terms
#6x^3 - 3x^2 = 30#

4) Divide all the terms on both sides by #3# to reduce the terms
#2x^3 - x^2 = 10#

5) Factor out #x^2#
#x^2 (2x - 1) = 10#