Question

Help!! idk how to solve this

Answer 1

Start by factoring the two denominators.

Answer 2

for the first question?

Answer 3

I only see one question which is the multiplication of two fractions. Each fraction has a numerator and a denominator. The numerators are simple expressions that can't be factored. The two denominators are quadratics that can be factored.

Rewrite everything, but with the two denominators factored.

Answer 4

This is your problem, right?

\(\dfrac{x+4}{x^2 + 5x + 6} \cdot \dfrac{x + 3}{x^2 - 16} \)

Answer 5

yes it is

Answer 6

Rewrite the entire thing, except factor the denominators:

\(\dfrac{x+4}{\color{red}{x^2 + 5x + 6}} \cdot \dfrac{x + 3}{\color{green}{x^2 - 16}} \)

\(=\dfrac{x+4}{\color{red}{(~~~~~~~~~)(~~~~~~~~~)}} \cdot \dfrac{x + 3}{\color{green}{(~~~~~~~~~)(~~~~~~~~~)}} \)

I set up the next step.
Factor the left denominator (in red) and write inside the red parentheses.
Factor the right denominator (in green) and write it inside the green parentheses.

Answer 7

red: (x+3)(x+2)
green: (x-4)(x+4) .... right?

Answer 8

Excellent.

Answer 9

Now you have this:

\(=\dfrac{x+4}{(x + 3)(x + 2)} \cdot \dfrac{x + 3}{(x-4)(x+4)}\)

When you multiply fractions, you multiply the numerators together and the denominators together:

\(=\dfrac{(x+4)(x + 3)}{(x + 3)(x + 2)(x-4)(x+4)}\)

Now you can cancel out common terms in the numerator and denominator.

Answer 10

you get left with 1/(x+2)(x-4) ?

Answer 11

\(=\dfrac{\cancel{(x+4)}\cancel{(x + 3)}~~1}{\cancel{(x + 3)}(x + 2)(x-4)\cancel{(x+4)}}\)

\(=\dfrac{1}{(x + 2)(x-4)}\)

You are correct. Good job!

Answer 12

thank you! could you help me with one more question? @mathstudent55