How do you factor completely #5x^3 - 45x#?

1 Answer
Apr 26, 2016

Answer:

#5x^3-45x = 5x(x-3)(x+3)#

Explanation:

Notice that both of the terms are divisible by #5x#, so we can separate that out as a factor first, then use the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=x# and #b=3# as follows:

#5x^3-45x = 5x(x^2-9) = 5x(x^2-3^2) = 5x(x-3)(x+3)#