Respuesta :
Answer
1 / (x + 3)(x + 4)
That is, 1 over the quantity x plus 3 times the quantity x minus 4.
Step by step explanation
The given expression = [tex]\frac{x-4}{x^2 - x - 12} . \frac{x-4}{x^2 - 8x + 16}[/tex]
Let's factor the denominators.
= [tex]\frac{x-4}{(x - 4)(x + 3)} .\frac{x-4}{(x - 4)(x - 4)}[/tex]
Cancelling (x- 4) both in the numerator and the denominator, we get
= [tex]\frac{1}{(x + 3)} .\frac{1}{(x- 4)}[/tex]
= 1/(x + 3)(x - 4)
The answer is 1 over the quantity x plus 3 times the quantity x minus 4.
Thank you.
Answer:
[tex]\frac{1}{(x+3)(x-4)}[/tex]
Step-by-step explanation:
The expression we are given is
[tex]\frac{x-4}{x^2-x-12}\times \frac{x-4}{x^2-8x+16}[/tex]
We factor the denominator of each expression.
To factor
x²-x-12, we find factors of -12 that sum to -1.
-4(3) = -12 and -4+3 = -1; this gives us
[tex]\frac{x-4}{(x-4)(x+3)}[/tex]
Cancelling the common factors in the numerator and denominator, we have
[tex]\frac{1}{x+3}[/tex]
To factor
x²-8x+16, we find factors of 16 that sum to -8;
-4(-4) = 16 and -4+-4 = -8. This gives us
[tex]\frac{x-4}{(x-4)(x-4)}[/tex]
Cancelling the common factors in the numerator and denominator, we have
[tex]\frac{1}{x-4}[/tex]
This gives us
[tex]\frac{1}{x+3}\times \frac{1}{x-4}\\\\=\frac{1}{(x+3)(x-4)}[/tex]
