## SIMPLIFYING RATIONAL EXPRESSIONS WORKSHEET

Simplify the following rational expression into their lowest forms.

(i)  (6x2+9x)/(3x2-12x)

(ii)  (x2+1)/(x4-1)

(iii)  (x3-1)/(x2+x+1)

(iv)  (x3-27)/(x2-9)

(v)  (x4+x2+1)/(x2+x+1)

(vi)  (x3+8)/(x4+4x2+16)

(vii)  (2x2+x-3)/(2x2+5x+3)

(viii)  (2x4-162)/(x2+9) (2x-6)

(ix)  [(x-3) (x2-5x+4)] / [(x-4) (x2-2x-3)]

(x)  [(x-8)(x2+5x-50)]/[(x+10)(x2-13x+40)]

(xi)  (4x2+9x+5)/(8x2+6x-5)

(i)  Solution :

(6x2+9x)/(3x2-12x)

Let f(x)  =  (6x2+9x)/(3x2-12x)

We may factor 3x from the numerator and denominator.

f(x)  =  3x(2x+3)/3x(x-4)

f(x)  =  (2x+3)/(x-4)

So, the value f(x) is (2x+3)/(x-4).

(ii)  Solution :

(x2+1)/(x4-1)

Let f(x)  =  (x2+1)/(x4-1)

f(x)  =  (x2+1)/((x2)2-(12)2)

Using the algebraic identity a2-b2, we may expand

a2-b2  =  (a+b)(a-b)

(x2)2-(12)2  =  (x2+1)(x2-1)

(x2)2-(12)2  =  (x2+1)(x+1)(x-1)

By applying factors in f(x), we get

f(x)  =  (x2+1)/(x2+1)(x+1)(x-1)

f(x)  =  1/(x+1)(x-1)

f(x)  =  1/(x2-1)

So, the value of f(x) is 1/(x2-1).

(iii)  Solution :

(x3-1)/(x2+x+1)

Let f(x)  =  (x3-1)/(x2+x+1)

Now we are going to use the following algebraic formula

a3-b3  =  (a-b) (a2+ab+b2)

x3-13  =  (x-1)(x2+x+1)

By applying factors in f(x), we get

f(x)  =  (x-1)(x2+x+1)/(x2+x+1)

f(x)  =  (x-1)

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

(iv)  Solution :

(x3-27)/(x2-9)

Let f(x)  =  (x3-27)/(x2-9)

f(x)  =  (x3-33)/(x2-32)

Now we are going to use the following algebraic formula

a3-b3  =  (a-b) (a2+ab+b2)

a2-b2  =  (a+b) (a-b)

x3-3 =  (x-3)(x2+3x+9)

x2-3 =  (x+3) (x-3)

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

f(x)  =  [(x-3)(x2+3x+9)]/[(x+3) (x-3) ]

f(x)  =  (x2+3x+9)/(x+3)

So, the value of f(x) is (x2+3x+9)/(x+3).

(v)  Solution :

(x4+x2+1)/(x2+x+1)

Let f(x)  = (x4+x2+1)/(x2+x+1)

Now we are going to use the following algebraic formula

(x4+x2+1)  =  (x2+1)2 - x2

f(x)  =  [(x2+1)2-x2]/(x2+x+1)

By applying factors in f(x), we get

f(x)  =  [(x2+1+x) (x2+1-x)]/(x2+x+1)

f(x)  =  [(x2+x+1 ) (x2-x+1)]/(x2+x+1)

f(x)  =  (x2-x+1)

So, the value of f(x) is (x2-x+1).

(vi)  Solution :

(x3+8)/(x4+4x2+16)

Let f(x)  =  (x3+8)/(x4+4x2+16)

x3+8  =  x3+23

x3+23  =  (x+2)(x2+2x+4)

(x4+4x2+16)  =  (x2+4)2-(2x)2

(x4+4x2+16)  =  [(x2+4)+(2x)][(x2+4)-(2x)]

(x4+4x2+16)  =  (x2+2x+4)(x2-2x+4)

By applying the factors, we get

=  (x+2)(x2+2x+4)/(x2+2x+4)(x2-2x+4)

=  (x+2)/(x2-2x+4)

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

(vii)   Solution :

(2x2+x-3)/(2x2+5x+3)

Let f(x)  = (2x2+x-3)/(2x2+5x+3)

2x2+x-3  =  (x-1) (2x+3)

2x2+5x+3  =  (x+1)(2x+3)

By applying the factored form in f(x), we get

f(x)  =  (x-1) (2x+3)/(x+1)(2x+3)

f(x)  =  (x-1)/(x+1)

So, the value of f(x) is (x-1)/(x+1).

(viii)  Solution :

(2x4-162)/(x2+9) (2x-6)

Let f(x)  =  (2x4-162)/(x2+9) (2x-6)

2x4-162  =  2(x4-81)

=  2((x2)2-(92)2)

=  2(x2+9) (x2-9)

2x4-162  =  2(x2+9) (x2-32)

2x4-162  =  2(x2+9) (x+3)(x-3)

2x-6  =  2(x-3)

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

f(x)  =  2(x2+9) (x+3)(x-3)/(x2+9) 2(x-3)

f(x)  =  x+3

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

(ix)  Solution :

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

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

x2-5x+4  =  (x-1)(x-4)

x2-2x-3  =  (x-3)(x+1)

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

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

By simplifying, we get

f(x)  =  (x-1)/(x+1)

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

(x)  Solution :

[(x-8)(x2+5x-50)]/[(x+10)(x2-13x+40)]

Let f(x)  =  [(x-8)(x2+5x-50)]/[(x+10)(x2-13x+40)]

x2+5x-50  =  (x+10)(x-5)

(x2-13x+40)  =  (x-8)(x-5)

f(x)  =  [(x-8)(x+10)(x-5)]/[(x+10)(x-8)(x-5)]

f(x)  =  1

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

(xi)  Solution :

(4x2+9x+5)/(8x2+6x-5)

Let f(x)  =  (4x2+9x+5)/(8x2+6x-5)

4x2+9x+5  =  (x+1) (4x+5)

8x2+6x-5  =  (4x+5) (2x-1)

f(x)  = (x+1)/(2x-1)

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

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