How do you use the remainder theorem to see if the #p+5# is a factor of #p^4+6p^3+11p^2+29p-13#?

1 Answer
EZ as pi
Feb 8, 2017

Not a factor.

Explanation:

Before even using the remainder theorem, you can see that #(p+5)# will not be a factor because the expression ends with #-13# which is not divisible by 5.

But looking at it again using the remainder theorem...

Call The expression #f(p)#

#f(p) = p^4 +6p^3+11p^2+29p-13#

If #p+5 = 0 rarr p = -5#

Calculate #f(-5)# by substituting (-5) for every #p#

If the answer is equal to #0#, it means that #(p+5)# is a factor.

If the answer is not #0#, then the value you get will be the remainder if you divide by #(p+5)#

#f(p) = p^4 +6p^3+11p^2+29p-13#

#f(-5) = (-5)^4 +6(-5)^3+11(-5)^2+29(-5)-13#

#=625-750+275-145 -13#

#=-8#

Nope, not a factor!

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