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What are the domain restrictions of q^2−7q−8 divided by q^2+3q−4 ?



q≠1 and q≠−8

q≠−1 and q≠8

q≠−1 and q≠4

q≠1 and q≠−4

Respuesta :

kritter1
Factorize the polynomials individually; 

q²-7q-8= (q-8)(q+1) <-- denominator 
q²+3q-4= (q-1)(q+4) 

When a number is divided by 0, it becomes indefinite (creating a hole in the graph). 

(q-8)(q+1)=0 
q=8
q=-1

Therefore, the second option.

Hope I helped :) 

The domain restrictions of [tex]\frac{q^2-7q-8 }{q^2+3q-4}[/tex]  are q≠1 and q≠−4

The given expression is:

[tex]\frac{q^2-7q-8 }{q^2+3q-4}[/tex]

The expression will be invalid when the denominator equals 0

The numerator = [tex]q^2 - 7q - 8[/tex]

The denominator = [tex]q^2 + 3q - 4[/tex]

Equate the denominator to zero

[tex]q^2 + 3q - 4 = 0[/tex]

Solve for q in the equation [tex]q^2 + 3q - 4 = 0[/tex] by factorization

[tex]q^2 + 3q - 4 = 0\\q^2 -q + 4q - 4 = 0\\q(q - 1) + 4(q - 1) = 0\\(q-1)(q+4) = 0\\q - 1 = 0\\q = 1\\q + 4 = 0\\q = -4[/tex]

For the rational expression to be valid [tex]q \neq 1[/tex] and [tex]q \neq -4[/tex]

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