Respuesta :

MrGumby
h(x) = x - 5

However, it had a domain of all real numbers such that x does not equal 3, since that would cause you to divide by 0. 
f(x)=x² - 8x + 15,
15=(-3)*(-5), -3+(-5)=-8
f(x)=x² - 8x + 15 = (x-3)(x-5)

g(x)=x-3

h(x)=f(x):g(x)

[tex]h(x)= \frac{(x-3)(x-5)}{x-3} = x-5, h(x)=x-5 Denominator cannot be equal 0. x-3=0, x=3 [/tex]

Domain of h(x) is (-∞,3) U (3, +∞)

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