How do you simplify #(p+4)/(p^2+6p+8)#?
2 Answers
Explanation:
Explanation:
Step A:

It's in the form "
#ap^2+bp+c# ," which means your factors#p_1# and#p_2# must add to be#b# and multiply to be#a\cdotc# .
In other words:#\bb(p_1+p_2=b)# and#\bb(p_1(p_2)=a\cdotc)# 
#a=1# ,#b=6# ,#c=8# ; so#p_1+p_2=6# and#p_1(p_2)=1\cdot8=8#
The only two factors that fulfill these requirements are#\bb2# and#\bb4# . Which means you can factor the polynomial#p^2+6p+8# to#(p+2)(p+4)# . 
Your new expression is
#(p+4)/((p+2)(p+4))# .
See anything you can cross out?
Step B:

We see that
#p+4# occurs twice, one in the numerator, and one in the denominator.
#(\color(red)(p+4))/((p+2)(\color(red)(p+4))# 
You can cancel out similar terms, so we cross out those two.
#\cancel(\color(red)(p+4))/((p+2)(\cancel(\color(red)(p+4)))# 
Remove the crossed out parts...
#1/((p+2)(1))#
Step C: