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) = [(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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