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2 answers:
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Answer:
1. =
2. =
3. =
4. =
Step-by-step explanation:
1.
Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)
Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.
2.
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)
Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)
Putting factors
Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.
3.
Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)
Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)
Putting factors
Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)
Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.
4.
Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)
Converting ÷ sign into multiplication we will take reciprocal of the second term
Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.