How do you show that x+1 is a factor of #x^4+2x^3-2x^2-6x-3#?

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Answer 1 · 2017-01-07T09:31:55

See explanation.

If #P_n(x) - a_0x^n+a_1x^(n-1)+...+a_(n-1)x+a_n#;

the sum of the coefficients

#P_n(1) = a_0+a_1+a_2+...+a_(n-1)+a_n=0 #

shows that 1 is a zero of #P_n. and so,

(x-1) is a factor of # P_n(x)#, Likewise, if

#P_(n-1) =(-1)^n(a_0-a_1+a_2-...+(-1)^a_n=0#, then

#-1# is a zero of #P_n(x), and so,

#(x-(-1)=(x+1)# is a factor.

Here, #P_4(-1)=1-2-2+6-3=2#, and so, ( x+1) is a factor.

Answer 2 · 2017-01-07T10:22:04

By using the factor theorem

#(x-a) # is a factor of #P(x)=> P(a)=0#

proof

Let #P(x)# be divided by #(x-a) # giving quotient #Q(x)#and remainder #R#

then

#P(x)=(x-a)Q(x)+R#

#P(a)=cancel((a-a)Q(x))+R#

#R=0=>(x-a)# is a factor of #P(x)#

in this case

to show #(x+1) #is a factor of

#P(x)=x^4+2x^3-2x^2-6x-3#

#a=-1#

#P(-1)=(-1)^4+2xx(-1)^3-2xx(-1)^2-6xx(-1)-3#

#P(-1)=1-2-2+6-3=-3+6-3=0#

#P(-1)=0=>(x+1)# is a a factor #P(x)#