# Division of Algebraic Expressions

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 doing division of 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 – 5

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

2x + 1
2x + 3   4x^2 + 8x – 5
−  4x^2 + 6x

2x – 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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