# Division of Algebraic Expressions

1. Division Examples
2. Algebraic Long Division

Division of algebraic expressions often requires cancelling out terms and exponents, as we look to simplify the terms involved.

Examples

(1.1)

Simplify    {\frac{5x^4 \space + \space 11x^2}{2x}}.

Solution

As both terms above share the same denominator, we can split this into 2 fractions.

{\frac{5x^4}{2x}} + {\frac{11x^2}{2x}}       From here we can cancel as required in both fractions.

{\frac{5 \times x \times x \times {\cancel{x}} \times {\cancel{x}}}{2 \times {\cancel{x}} \times {\cancel{x}}}} + {\frac{11 \times {\cancel{x}} \times x}{2 \times {\cancel{x}}}}

{\frac{5x^2}{2}} + {\frac{11x}{2}}       Now we can put the 2 fractions together for the full simplified expression.

{\frac{5x^2 \space + \space 11x}{2}}

With enough practice of dividing with algebraic expressions, one is usually able to do the cancelling of terms in a fraction in their head without the need to write the full terms out as we did here.

(1.2)

Simplify    {\frac{4x \space {\text{--}} \space 2x^4}{x^3}}.

Solution

{\frac{4x}{x^3}}{\frac{2x^4}{x^3}}

=   {\frac{4}{x^2}}{\frac{2x^3}{x^2}}   =   {\frac{4 \space {\text{--}} \space 2x^3}{x^2}}

(1.3)

Simplify    6a^{2}b^{5} \space \div \space 2ab^{6}.

Solution

When presented as a division sum, we can still write the problem out as a fraction.

{\frac{6a^{2}b^{5}}{2ab^{6}}}

In this case, due to the commutative property of multiplication, we can write out as 3 separate fractions multiplied together. Then simplify accordingly.

Before putting the new fractions back together again.

{\frac{6}{2}} . {\frac{a^2}{a}} . {\frac{b^5}{b^6}}    =    {\frac{3}{1}} . {\frac{a}{1}} . {\frac{1}{b}}    =    {\frac{3a}{b}}

## Division of Algebraic Expressions,Long Division

Sometimes when dealing with division of algebraic expressions, algebraic long division is an approach that can be used.

Say we had a division sum,

( x^2 + 4x + 4 )  ÷  ( x + 2 ).

To do long division with these algebraic expressions, we set them up as we would with long division for numbers,

x + 2   x^2 + 4x + 4.

Then proceed following several specific steps, shown below.

Examples

(2.1)

( 2x^2 + 6x + 4 )  ÷  ( x + 1 )

Solution

x + 1   2x^2 + 6x + 4

1)      [  {\frac{2x^2}{x}} = 2x  ,   2x(x+1)  =  2x^2+2x  ]

2x
x + 1   2x^2 + 6x + 4
−  2x^2 + 2x

4x + 4

2)      [  {\frac{4x}{x}} = 4   ,   4(x+1)  =  4x+4  ]

2x + 4
x + 1   2x^2 + 6x + 4
−  2x^2 + 2x

4x + 4
−  4x + 4

0

A single number, even zero, is an expression with an exponent less than the original divisor.

As a single number like  0  can be thought of as the term  0x0.

=>   ( 2x^2 + 6x + 4 )  ÷  ( x + 1 )   =   2x + 4

(2.2)

( 4x^2 + 8x − 5 )  ÷  ( 2x + 3 )

Solution

2x + 3   4x^2 + 8x - 5

1)      [  {\frac{4x^2}{2x}} = 2x   ,   2x(2x+3)  =  4x^2+6x  ]

2x
2x + 3   4x^2 + 8x - 5
−  4x^2 + 6x

2x {\text{--}} 5

2)      [  {\frac{2x}{2x}} = 1   ,   1(2x+3)  =  2x+3  ]

2x + 1
2x + 3   4x^2 + 8x \space - \space 5
−  4x^2 + 6x

2x \space {\text{--}} \space 5
−  2x+3

– 8

This long division of algebraic expressions example has a remainder other than  0.
When this happens, we add or subtract this remainder depending on its nature, as a fraction over the original divisor.

( 4x^2 + 8x \space{\text{--}}\space 5 )  ÷  ( 2x + 3 )   =   2x + 1 \space {\text{--}} \space {\large{{\frac{8}{2x+3}}}}

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