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Is it possible to manipulate: $\frac{x^2 -2}{x+1}$ such that the numerator has a lower power than the denominator?

For example: $\frac{x+3}{x+2}$ would be $1+\frac{1}{x+2}$.

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5 Answers 5

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Hint $$\frac{x^2 -2}{x+1}=\frac{x^2 -1}{x+1}-\frac{1}{x+1}$$

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

In general, we can perform division of polynomial. When $x^2-2$ is divided by $x+1$, the quotient is $x-1$ and the remainder is $-1$.

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

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This is the result of the polynomial division between $x^2-2$ and $x+1$: $$ {x^2-2\over x+1}={(x^2+2x+1)-(2x+2)+(2-1-2)\over x+1}=x-1-{1\over x+1} $$

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This requires nothing smart: just do long division.

$$ \begin{array}{c|c} \begin{array}[t]{r} \hphantom{-}x^2+0x-2 \\ -x^2-\hphantom{0}x\hphantom{{}-0} \\ \hline -x-2\\ x+1\\ \hline -1 \end{array} & \begin{array}[t]{l} x+1\\[4px] \hline x-1 \end{array} \end{array} $$

Or with an abbreviated scheme: $$ \begin{array}{r|rr|r} & 1 & 0 & -2 \\ -1 & & -1 & 1 \\ \hline & 1 & -1 & -1 \end{array} $$ Both tell you that $$ x^2-2=(x-1)(x+1)-1 $$

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Yes, this is always possible when dividing one polynomial by another. Suppose we start with $$\frac{P(x)}{Q(x)}$$ where $P(x)=a_nx^n+$ lower-order terms, and $Q(x)=b_mx^m+$ lower-order terms, with $n \ge m$. Then we have $$\frac{P(x)}{Q(x)}=\frac{a_n}{b_m}x^{n-m}+\frac{R(x)}{Q(x)}$$ where $$R(x)=P(x)-\frac{a_n}{b_m}x^{n-m}Q(x)$$ is a polynomial of degree $\le n-1$ (because its $n^\text{th}$-order coefficient is $a_n-\frac{a_n}{b_m}b_m=0$).

You can repeat this process with $R(x)/Q(x)$, until you end up with $$\frac{P(x)}{Q(x)}=S(x)+\frac{T(x)}{Q(x)}$$ where $S$ is a polynomial of degree $n-m$ and $T$ is a polynomial of degree $\le m-1$.

In practice, this process is very simple. At each step, you just look at the highest-order terms in the fraction, and the next term of $S$ is their quotient.

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