SIMPLIFYING POLYNOMIAL EXPRESSIONS IN FRACTIONS

Simplifying polynomial expressions is nothing but expressing the the rational expression to lowest term or simplest form.

The following steps ill be useful to simple rational expressions. 

Step 1 :

Factor both numerator and denominator, if it is possible.

Step 2 :

Identify the common factors in both numerator and denominator. 

Step 3 :

Remove the common factors found in both numerator and denominator.

Example 1 :

[(x2-2x)/(x+2)]  [(3x+6)/(x-2)]

Solution :

Let f(x)  =   [(x2-2x)/(x+2)]  [(3x+6)/(x-2)]

f(x)  =  [(x2-2x)/(x+2)]  [(3x+6)/(x-2)]

f(x)  =  [x(x-2)/(x+2)]  [3(x+2)/(x-2)]

f(x)  =  3x

So, the value of f(x) is 3x.

Example 2 :

[(x2-81)/(x2-4)]  [(x2+6x+8)/(x2-5x-36)]

Solution :

Let f(x)  =  [(x2-81)/(x2-4)]  [(x2+6x+8)/(x2-5x-36)]

x- 81  =  x2- 92  ==>  (x+9)(x-9)

x- 4  =  x2- 22  ==> (x+2)(x-2)

x2+6x+8  =  (x+2)(x+4)

x2-5x-36  =  (x-9)(x+4)

f(x)  =  [(x+9)(x-9)/(x+2)(x-2)] ⋅ [(x+2)(x+4)/(x-9)(x+4)]

By simplifying (x+9)/(x-2)

So, the value of f(x) is (x+9)/(x-2).

Example 3 :

[(x2-3x-10)/(x2-x-20)]  [(x2-2x+4)/(x3+8)]

Solution :

Let f(x)  =  [(x2-3x-10)/(x2-x-20)]  [(x2-2x+4)/(x3+8)]

x2-3x-10  =  (x-5)(x+2)

x2-x-20  =  (x-5)(x+4)

a3+b =  (a+b)(a2-ab+b2)

x3+2 =  (x+2)(x2-2x+4)

By applying the factors in f(x), we get

 =  [(x-5)(x+2)/(x-5)(x+4) [(x2-2x+4)/(x+2)(x2-2x+4)]

=  1/(x+4)

So, the value of f(x) is 1/(x+4)

Example 4 :

[(x2-16)/(x2-3x+2)]  [(x2-4)/(x3+64)]  

[(x2-4x+16)/(x2-2x-8)]

Solution :

Let f(x)  =  [(x2-16)/(x2-3x+2)]  [(x2-4)/(x3+64)]  

[(x2-4x+16)/(x2-2x-8)]e

x2-16  =  x2-4 ==>  (x+4)(x-4)

x2-3x+2  =  (x-1)(x-2)

x2-4  =  x2-22  ==>  (x+2)(x-2)

x3+64  =  x3+4 ==> (x+4)(x2-4x+16) 

x2-2x-8  =  (x-4)(x+2)

=  [(x+4)(x-4)/(x-1)(x-2)][(x+2)(x-2)/(x+4)(x2-4x+16)]

[(x2-4x+16)/(x-4)(x+2)]

f(x)  =  1/(x-1)

So, the value of f(x) is 1/(x-1).

Example 5 :

[(3x2+2x-1)/(x2-x-2)] [(2x2-3x-2)/(3x2+5x-2)]

Solution :

Let f(x)  =  [(3x2+2x-1)/(x2-x-2)]

 [(2x2-3x-2)/(3x2+5x-2)]

(3x2+2x-1)  =  (3x-1) (x+1)

(x2-x-2)  =  (x-2) (x+1)

(2x2-3x-2)  =  (2x+1) (x-2)

(3x2+5x-2) =   (2x-1) (x+2)

By applying the factors in f(x), we get

=  [(3x-1)(x+1)/(x-2) (x+1)][(2x+1) (x-2)/(2x-1) (x+2)]

=  (2x+1)/(x+2)

So, the value of f(x) is (2x+1)/(x+2).

Example 6 :

[(2x-1)/(x2+2x+4)] [(x4-8x)/(2x2+5x-3)] 

[(x+3)/(x2-2x)]

Solution :

Let f(x)  =  [(2x-1)/(x2+2x+4)] [(x4-8x)/(2x2+5x-3)] 

[(x+3)/(x2-2x)]

x4-8x  =  x(x3-23)

x4-8x  =  x(x-2)(x2+2x+4)

2x2+5x-3  =  (2x-1)(x+3)

x2-2x  =  x(x-2)

By applying the factors in f(x), we get

=  [(2x-1)/(x2+2x+4)][x(x-2)(x2+2x+4)/(2x-1)(x+3)] 

[(x+3)/x(x-2)]

=  1

So, the value of f(x) is 1.

Example 7 :

[(a+b)/(a-b)] [(a3-b3)/(a3+b3)]

Solution :

Let f(x)  =  [(a+b)/(a-b)] [(a3-b3)/(a3+b3)]

=  [(a+b)/(a-b)][(a-b)(a2+ab+b2)/(a+b) (a2-ab+b2)]

=  (a2+ab+b2)/(a2-ab+b2)

So, the value of f(x) is (a2+ab+b2)/(a2-ab+b2).

Example 8 :

[(x2-9y2)/(3x-3y)]  [(x2-y2)/(x2+4xy+3y2)]

Solution :

Let f(x)  =  [(x2-9y2)/(3x-3y)]  [(x2-y2)/(x2+4xy+3y2)]

x2-9y=  x2-(3y)2

x2-9y2  =  (x+3y)(x-3y)

3x-3y  =  3(x-y)

x2-y=  (x+y)(x-y)

x2+4xy+3y=  (x+3y)(x+y)

By applying the factors in f(x), we get

=  [(x+3y)(x-3y)/3(x-y)]⋅[(x+y)(x-y)/(x+3y)(x+y)]

By simplifying, we get

=  (x-3y)/3

So, the value of f(x) is (x-3y)/3.

Example 9 :

[(x2-4x-12)/(x2-3x-18)]  [(x2-2x-3)/(x2+3x+2)]

Solution :

Let f(x)  =  [(x2-4x-12)/(x2-3x-18)] 

 [(x2-2x-3)/(x2+3x+2)]

x2-4x-12  =  (x-6)(x+2)

x2-3x-18  =  (x-6)(x+3)

x2-2x-3  =  (x-3)(x+1)

x2+3x+2  =  (x+1)(x+2)

f(x)  =  [(x-6)(x+2)/(x-6)(x+3)]⋅[(x-3)(x+1)/(x+1)(x+2)]

f(x)  =  (x-3)/(x+3)

So, the value of f(x) is (x-3)/(x+3).

Example 10 :

[(x2-3x-10)/(x2-x-20)][(x2-4x+16)/(x3+64)]

Solution :

Let f(x)  =  [(x2-3x-10)/(x2-x-20)][(x2-4x+16)/(x3+64)]

x2-3x-10  =  (x-5)(x+2)

x2-x-20  =  (x-5)(x+4)

x3+43  =  (x+4)(x2-4x+16)

By applying the factors in f(x), we get

f(x)  =  [(x-5)(x+2)/(x-5)(x+4)][(x2-4x+16)/(x+4)(x2-4x+16)]

f(x)  =  (x+2)/(x+4)2

So, the value of f(x) is (x+2)/(x+4)2.

Example 11 :

 [(x2-16)/(x-2)] [(x2-4)/(x3+64)]

Solution :

Let f(x)  =  [(x2-16)/(x-2)] [(x2-4)/(x3+64)]

x2-16  =  x2-42  ==>  (x+4)(x-4)

x2-4  =  x2-22  ==>  (x+2)(x-2)

x3+64  =  x3+43  ==>  (x+4)(x2-4x+16)

f(x)  =  [(x+4)(x-4)/(x-2)] [(x+2)(x-2)/(x+4)(x2-4x+16)]

f(x)  =  (x-4)(x-2)/(x2-4x+16)

So, the value of f(x) is (x-4)(x-2)/(x2-4x+16).

Example 12 :

[(x + 7)/(x2+14x+49)] [(x2+8x+7)/(x+1)]

Solution :

Let f(x)  =  [(x + 7)/(x2+14x+49)] [(x2+8x+7)/(x+1)]

x2+14x+49  =  (x+7)(x+7)

x2+8x+7  =  (x+1)(x+7)

By applying the factors in f(x), we get

f(x)  =  [(x+7)/(x+7)(x+7)] [(x+1)(x+7)/(x+1)]

f(x)  =  1

So, the value of f(x) is 1.

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