SIMPLIFYING POLYNOMIAL EXPRESSIONS IN FRACTIONS

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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)]

x2 - 81 = x2- 92  ==> (x + 9)(x - 9)

x2 - 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 + b3 = (a + b)(a2 - ab + b2)

x3 + 23 = (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 - 42  ==> (x + 4)(x - 4)

x2 - 3x + 2 = (x - 1)(x - 2)

x2 - 4 = x2 - 22  ==> (x + 2)(x - 2)

x3 + 64  = x3 + 43  ==> (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)]

x2-9y2  =  x2-(3y)2

x2-9y2  =  (x+3y)(x-3y)

3x-3y  =  3(x-y)

x2-y2  =  (x+y)(x-y)

x2+4xy+3y2  =  (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)  =  [(x 2 - 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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