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Question

Show that #x+3# is a factor of #P(x)=x^4+3x^3-9x^2-27x#

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Answer 1 · 2018-03-26T12:01:16

#color(red)( x+3# #color(red)("is a factor of"# #color(red)(x^4+3x^3-9x^2-27x#

If #x+3# is a factor of #x^4+3x^3-9x^2-27x#,
then #x+3=0#

#therefore x=-3#

#(-3)^4+3(-3)^3-9(-3)^2-27(-3)#

#=cancel81-cancel81+cancel81-cancel81#

#=0#

#therefore# #color(red)( x+3# #color(red)("is a factor of"# #color(red)(x^4+3x^3-9x^2-27x#

~Hope this helps! :)

Answer 2 · 2018-03-26T12:01:38

Using the remainder theorem:

#f(x)=g(x)(x-a)+r#

Where #r# is the remainder and #g(x)# is the quotient.

We know from this that if we can make #(x-a)=0#, then this gives the remainder. If #P(x)# is divisible by #(x+3)# then the remainder will be zero:

#P(-3)=g(x)(-3+3)+0#

#P(-3)=0#

#(-3)^4+3(-3)^3-9(-3)^2-27(-3)=0#

So #(x+3)# is a factor of #P(x)#