How do you simplify #(p+4)/(p^2+6p+8)#?

2 Answers
Apr 17, 2018

Answer:

#= (1/(p+2))#

Explanation:

#(p + 4) / (p^2 + 6p + 8)#
#= (p+4)/((p+2)(p+4))#
#= (1/(p+2))#

Apr 17, 2018

Answer:

#1/(p+2)#

Explanation:

Step A: #\bb(\text(Simplify the denominator.))#

  1. 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)#

  2. #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)#.

  3. Your new expression is #(p+4)/((p+2)(p+4))#.
    See anything you can cross out?

Step B: #\bb(\text(Identify similar terms to cancel.))#

  1. 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))#

  2. You can cancel out similar terms, so we cross out those two.
    #\cancel(\color(red)(p+4))/((p+2)(\cancel(\color(red)(p+4)))#

  3. Remove the crossed out parts...
    #1/((p+2)(1))#

Step C: #\text(Remove the 1s, and)\bb(\text( you have your answer!))#
#1/(p+2)#