How do you factor #x^2+16x+48#?

1 Answer
Apr 12, 2018

Answer:

#(x+12)(x+4)=0#

Explanation:

the form that your equations is in is
#ax^2+bx+c#
Multiply a and c (1*48) to get 48. Now, what two numbers do you multiply to get 48 and add to get 16, which is b in your equation? The answer is 4 and 12. Because they are positive, put them as factors by putting "x+" in front of each and setting them as factors in an equation that equal 0.