How do you factor #n^ { 2} - 6n - 24= 0#?
1 Answer
Sep 16, 2017
Explanation:
This can be factored by completing the square and using the difference of squares identity:
#A^2-B^2 = (A-B)(A+B)#
with
#n^2-6n-24 = n^2-6n+9-33#
#color(white)(n^2-6n-24) = (n-3)^2-(sqrt(33))^2#
#color(white)(n^2-6n-24) = ((n-3)-sqrt(33))((n-3)+sqrt(33))#
#color(white)(n^2-6n-24) = (n-3-sqrt(33))(n-3+sqrt(33))#
So the factored form of the original equation can be written:
#(n-3-sqrt(33))(n-3+sqrt(33)) = 0#