How do you factor #3j^3-51j^2+210j#?

1 Answer
Apr 12, 2016

Answer:

f(j) = 3j(j - 7)(j - 10)

Explanation:

#f(j) = 3j(y) = 3j(j^2 - 17j + 70)#
Factor y by the new AC Method (Socratic Search).
Find two numbers knowing sum (-17= b) and product (c = 70).
These numbers have same sing because ac > 0
Factor pairs of (70) --> (7, 10)(-7, -10). This sum is (-17 = b).
Therefor,
y = (j - 7)(j - 10)
f(j) = 3j(j - 7)(j - 10)