How do you find the remainder for #(x^4-x^3-3x^2+4x+2)div(x+2)#?

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Answer 1 · 2017-11-16T12:38:00

The remainder is #=6#

Apply the remainder theorem

When we divide a polynomial #f(x)# by #(x-c)#

#f(x)=(x-c)q(x)+r#

#f(c)=0+r#

Here,

#f(x)=x^4-x^3-3x^2+4x+2#

and #(x-c)# is #(x-(-2))#

Therefore,

#f(-2)=(-2)^4-(-2)^3-3(-2)^2+4(-2)+2#

#=16+8-12-8+2#

#=6#

The remainder is #=6#