# Operations with algebraic fractions

## Sum and subtraction

In order to compute the sum or subtraction of algebraic fractions, first we have to convert the fractions to a common denominator, and then we can compute the sum or subtraction as if it was a fraction.

Once the denominator is computed, the numerator is just the sum or subtraction of the different terms.

Compute the sum of the following algebraic fractions $$\dfrac{x-1}{x+4} \ \mbox{and} \ \dfrac{x^2+2}{x+4}$$

In this case, both fractions have the same denominator, and we can then compute straight away: $$\dfrac{x-1}{x+4}+\dfrac{x^2+2}{x+4}=\dfrac{x-1+(x^2+2)}{x+4}=\dfrac{x^2+x+1}{x+4}$$\$

Compute the subtraction of the following algebraic fractions $$\dfrac{x^2+1}{x-2} \ \mbox{and} \ \dfrac{x+1}{x-1}$$

First, we have to convert the algebraic fractions into fractions with common denominator:

$$lcm\{x-2,x-1\}=(x-2)\cdot(x-1)$$

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

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

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

Now we compute:

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

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

Compute the subtraction of the following algebraic fractions $$\dfrac{x-2}{x+3} \ \mbox{and} \ \dfrac{x-1}{(x+1)^2}$$

First, we have to convert the algebraic fractions into fractions with common denominator:

$$lcm\{x+3,(x+1)^2\}=(x+3)\cdot(x+1)^2$$

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

$$=x\cdot(x^2+2x+1)+1\cdot(x^2+2x+1)=x^3+3x^2+3x+1 \Rightarrow$$

$$\Rightarrow \dfrac{x^3+3x^2+3x+1}{(x+3)\cdot(x+1)^2}$$

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

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

Now we compute:

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

$$=\dfrac{x^3+2x^2-x+4}{(x+3)(x+1)^2}$$

## Product

To compute the product of two algebraic fractions, the numerator of the product will be the numerators' product and the denominator of the product will be the denominators' product.

Compute the product of the following algebraic fractions $$\dfrac{x-1}{x+4}$$ and $$\dfrac{x^2+2}{x-2}$$.

We multiply numerators and denominators, and obtain the desired result:

$$\dfrac{x-1}{x+4}\cdot\dfrac{x^2+2}{x-2}=\dfrac{(x-1)\cdot(x^2+2)}{(x+4)\cdot(x-2)}=\dfrac{x\cdot(x^2+2)-1\cdot(x^2+2)}{x\cdot(x-2)+4\cdot(x-2)}=$$

$$=\dfrac{x^3-x^2+2x-2}{x^2+2x-8}$$

Compute the product of the following algebraic fractions $$\dfrac{x+5}{x}$$ and $$\dfrac{x^2-1}{x+3}$$.

We multiply numerators and denominators, and obtain the desired result:

$$\dfrac{x+5}{x}\cdot\dfrac{x^2-1}{x+3}=\dfrac{(x+5)\cdot(x^2-1)}{x\cdot(x+3)}=\dfrac{x\cdot(x^2-1)+5\cdot(x^2-1)}{x\cdot(x+3)}=$$

$$=\dfrac{x^3+5x^2-x-5}{x^2+3x}$$

## Division

In order to compute the division of two algebraic fractions, it is enough to multiply the algebraic fraction of the dividend by the inverted algebraic fraction of the divisor, that is, the numerator instead of the denominator, and vice versa.

Compute the division of the following algebraic fractions $$\dfrac{x-1}{x+4}$$ and $$\dfrac{x^2+2}{x-2}$$.

We multiply the first fraction by the second inverted, and obtain the desired result:

$$\dfrac{x-1}{x+4}\cdot\dfrac{x-2}{x^2+2}=\dfrac{(x-1)\cdot(x-2)}{(x+4)\cdot(x^2+2)}=\dfrac{x\cdot(x-2)-1\cdot(x-2)}{x\cdot(x^2+2)+4\cdot(x^2+2)}=$$

$$=\dfrac{x^2-3x+2}{x^3+4x^2+2x+8}$$

Compute the division of the following algebraic fractions $$\dfrac{x+5}{x}$$ and $$\dfrac{x^2-1}{x+3}$$.

We multiply the first fraction by the second inverted, and obtain the desired result:

$$\dfrac{x+5}{x}\cdot\dfrac{x+3}{x^2-1}=\dfrac{(x+5)\cdot(x+3)}{x\cdot(x^2-1)}=\dfrac{x\cdot(x+3)+5\cdot(x+3)}{x\cdot(x^2-1)}=$$

$$=\dfrac{x^2+8x+15}{x^3-x}$$