How do you factor #n^ { 2} - 6n - 24= 0#?

1 Answer
George C.
Sep 16, 2017

#(n-3-sqrt(33))(n-3+sqrt(33)) = 0#

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 #A=(n-3)# and #B=sqrt(33)# as follows:

#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#

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