How do you use the remainder theorem to determine whether #x+2# is a factor of #f(x)=3x^3-2x^2-6x-2#?

2 Answers
Jun 7, 2017

Not a factor.

Explanation:

According to the remainder theorem, #x - a# is a factor of #f(x) iff f(a) = 0#.

We are asked to determine whether #x + 2# is a factor of the given function, so let's evaluate #f(- 2)#:

#Rightarrow f(- 2) = 3 (- 2)^(3) - 2 (- 2)^(2) - 6 (- 2) - 2#

#Rightarrow f(- 2) = 3 (- 8) - 2 (4) + 12 - 2#

#Rightarrow f(- 2) = - 24 - 8 + 10#

#Rightarrow f(- 2) = - 22#

#therefore f(- 2) ne 0#

Therefore, #x + 2# is not a factor of #f(x)#.

Jun 7, 2017

All polynomials #P(x)# can be rewritten in terms of three other polynomials #Q(x)# (quotient), #D(x)# (divisor), and #R(x)# (remainder):
#(P(x))/(D(x))=Q(x)+(R(x))/(D(x))#.

This can be rewritten to #P(x)=Q(x)D(x)+R(x)#. Now, suppose that there is a constant #k# such that #D(k)=0#. Then, it follows that #P(k)=R(k)#.

This is the Remainder Theorem: if a polynomial #P(x)# is divided by #D(x)# with #D(k)=0#, then the remainder #R(x)# satisfies #R(k)=P(k)#.

If #x+2# is a factor of #3x^3-2x^2-6x-2#, then the remainder of #3x^3-2x^2-6x-2# divided by #x+2# must be #0#.

According to the remainder theorem, the remainder #R# (which is a constant number in this case as the divisor is linear) satisfies #R=P(k)#, where #k# is the solution to #x+2=0#, i.e. #-2#. Thus, #R=3(-2)^3-2(-2)^2-6(-2)-2=-22#. Since #R!=0#, #x+2# is not a factor of #3x^3-2x^2-6x-2#.