How do you divide #( x^4-2x^3-x)/(x(x+4))#?

1 Answer

#(x^4-2x^3-x)/(x(x+4))=(x^3-2x^2-1)/(x+4)=color(red)(x^2-6x+24-97/(x+4))#

Explanation:

From the given #(x^4-2x^3-x)/(x(x+4))#

We can reduce the degree of the dividend and the divisor

#(x^4-2x^3-x)/(x(x+4))#

factor the common monomial x

#(x(x^3-2x^2-1))/(x(x+4))#

#(cancelx(x^3-2x^2-1))/(cancelx(x+4))#

#(x^3-2x^2-1)/(x+4)#

Perform long division
#" " " " ""underline(x^2-6x+24" " " " " ")#
#x+4|~x^3-2x^2+0*x-1#
#"" " " " " underline(x^3+4x^2" " " " "" " " " "" " " " ")#
#" " " " " " " "-6x^2+0*x-1#
#" " " " " " " "underline(-6x^2-24x" " " " "" " " " ")#
#" " " " " " " " " " " " " " "24x-1#
#" " " " " " " " " " " " " " "underline(24x+96)#
#" " " " " " " " " " " " " " " " " " -97##larr#remainder

We write our answer this way

#("Dividend")/("Divisor")="Quotient"+("Remainder")/("Divisor")#

#(x^4-2x^3-x)/(x(x+4))=x^2-6x+24-97/(x+4)#

God bless....I hope the explanation is useful.