Respuesta :
We want to multiply [tex]\displaystyle{ \frac{r^2+7r+10}{3} [/tex] by [tex]\displaystyle{ \frac{3r-30}{r^2-5r-50} [/tex], and simplify.
The first thing we do in this type of problems, is factorize as much as possible, and see what cancels out.
[tex]\displaystyle{ r^2+7r+10[/tex] can be factored into (r+2)(r+5) as 2 and 5 are two numbers whose sum is 7, and product is 10.
Similarly [tex]\displaystyle{ r^2-5r-50[/tex] can be factoredinto (r-10)(r+5) as -10 and 5 are two numbers whose sum is -5, and product is -50.
Last, (3r-30) can be factored as 3(r-10).
Thus the product is equal to:
[tex]\displaystyle{ \frac{(r+2)(r+5)}{3}\cdot \frac{3(r-10)}{(r-10)(r+5)} [/tex],
so canceling all equal terms in the numerator and denominator we are left with
r+2.
Answer: A) r+2
The first thing we do in this type of problems, is factorize as much as possible, and see what cancels out.
[tex]\displaystyle{ r^2+7r+10[/tex] can be factored into (r+2)(r+5) as 2 and 5 are two numbers whose sum is 7, and product is 10.
Similarly [tex]\displaystyle{ r^2-5r-50[/tex] can be factoredinto (r-10)(r+5) as -10 and 5 are two numbers whose sum is -5, and product is -50.
Last, (3r-30) can be factored as 3(r-10).
Thus the product is equal to:
[tex]\displaystyle{ \frac{(r+2)(r+5)}{3}\cdot \frac{3(r-10)}{(r-10)(r+5)} [/tex],
so canceling all equal terms in the numerator and denominator we are left with
r+2.
Answer: A) r+2
