Question

itle=" \frac{x^{3} }{x^{2} + 2x + 1 } " alt=" \frac{x^{3} }{x^{2} + 2x + 1 } " align="absmiddle" class="latex-formula">
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

$\dfrac{x^3}{x^2+2x+1}=x-\dfrac{2x^2+x}{x^2+2x+1}$

then

$\dfrac{x^3}{x^2+2x+1}=x-2+\dfrac{3x+2}{x^2+2x+1}$

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

$\dfrac{x^3}{(x+1)^2}=\dfrac{(x+1)^3-(3x^2+3x+1)}{(x+1)^2}$

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

$\dfrac{x^3}{(x+1)^2}=x+1-\dfrac{3x^2+3x+1}{(x+1)^2}$

Next,

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

so that

$\dfrac{x^3}{(x+1)^2}=x+1-\dfrac{3((x+1)^2-(2x+1))}{(x+1)^2}-\dfrac{3x+1}{(x+1)^2}$

$\dfrac{x^3}{(x+1)^2}=x+1-3+\dfrac{6x+3}{(x+1)^2}-\dfrac{3x+1}{(x+1)^2}$

$\dfrac{x^3}{(x+1)^2}=x-2+\dfrac{3x+2}{(x+1)^2}$

which is the same result as before.