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)
x3-33 = (x-3)(x2+3x+9)
x2-32 = (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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