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.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. 

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Finding Derivative of a Function

    Mar 29, 24 12:11 AM

    Problems on Finding Derivative of a Function

    Read More

  2. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  3. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More