Respuesta :
x-2 over x2+x-6 divided x2+5x+4 over x+4 requires 2 factoring, canceling like factors, and inverting the divisor and multiplying.
(x-2) over (x+3)(x-2) divided by (x+4)(x+1) over (x+4) after factoring twice
canceling like factors
1 over (x+3) divided by (x+1) over 1
invert denominator and multiply
answer 1 over (x+3)(x+1)
(x-2) over (x+3)(x-2) divided by (x+4)(x+1) over (x+4) after factoring twice
canceling like factors
1 over (x+3) divided by (x+1) over 1
invert denominator and multiply
answer 1 over (x+3)(x+1)
Answer:
[tex]1/((x+3)*(x+1))[/tex]
Step-by-step explanation:
We can write the expression as:
[tex]((x-2)/(x^2+x-6))/((x^2+5*x+4)/(x+4))[/tex]
We can write the term as a single division by multiplying the terms in numerator by the inverse on the terms in the denominator:
[tex]((x-2)/(x^2+x-6))*((x+4)/(x^2+5*x+4)/)[/tex]
[tex]((x-2)*(x+4))/((x^2+x-6)*(x^2+5*x+4))[/tex]
We can simplfy the denominator terms as:
[tex]x^2+x-6=(x+3)*(x-2)[/tex]
[tex]x^2+5*x+4=(x+4)*(x+1)[/tex]
Therefore (x-2) and (x+4) in the numerator and denominator cancels out and we are left with:
[tex]1/((x+3)*(x+1))[/tex]