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(x^(3) )/(x^(2) + 2x + 1 )
How do I divide a monomial by a polynomial? ​

User Sgjesse
by
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1 Answer

4 votes


x^3=\boxed{x}\cdot x^2, and


\boxed{x}(x^2+2x+1)=x^3+2x^2+x

Subtract this from
x^3 to get a remainder of


x^3-(x^3+2x^2+x)=-2x^2-x


-2x^2=\boxed{-2}\cdot x^2, and


\boxed{-2}(x^2+2x+1)=-2x^2-4x-2

Subtract this from the previous remainder to get a new remainder of


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


3x does not divide
x^2, so we stop here.

What we've done is to write


(x^3)/(x^2+2x+1)=x-(2x^2+x)/(x^2+2x+1)

then


(x^3)/(x^2+2x+1)=x-2+(3x+2)/(x^2+2x+1)

and we stop here because the remainder term
(3x+2) has a degree less than the degree of the denominator.

Alternatively, we can be a bit tricky and notice that


x^2+2x+1=(x+1)^2

Now,


(x+1)^3=x^3+3x^2+3x+1

so that


(x^3)/((x+1)^2)=((x+1)^3-(3x^2+3x+1))/((x+1)^2)

We can divide the first term by
(x+1)^2 easily to get


(x^3)/((x+1)^2)=x+1-(3x^2+3x+1)/((x+1)^2)

Next,


(x+1)^2=x^2+2x+1

so that


(x^3)/((x+1)^2)=x+1-(3((x+1)^2-(2x+1)))/((x+1)^2)-(3x+1)/((x+1)^2)


(x^3)/((x+1)^2)=x+1-3+(6x+3)/((x+1)^2)-(3x+1)/((x+1)^2)


(x^3)/((x+1)^2)=x-2+(3x+2)/((x+1)^2)

which is the same result as before.

User Algorys
by
7.9k points

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