Question

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

Answer 1

`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.