## Jamierox4ev3r one year ago Check my work someone, not sure if I'm going about this in the correct manner

1. Jamierox4ev3r

Give me a second to type out the problem, thank you

2. Jamierox4ev3r

Simplify the rational expression: $$\Large\frac{10^{2}+29x+21}{x^{2}-25}\times\frac{x+5}{2x+3}$$

3. Jamierox4ev3r

So far, I've factored the top portion. So $$10x^{2}+29x+21$$ should turn into (5x+7)(2x+3) while $$x^{2}-25$$ should be (x-5)(x+5)

4. Jamierox4ev3r

*factored both the numerator and denominator of the first part of the expression

5. Jamierox4ev3r

So I tried cross-multiplying, but that brought me more complications...however, I noticed a few things $$\huge\frac{(5x-7)(2x+3)}{(x-5)(x+5)}\times\frac{x+5}{2x+3}$$ $$\huge\frac{(5x-7)\color{red}{(2x+3)}}{(x-5)\color{red}{(x+5)}}\times\frac{\color{red}{x+5}}{\color{red}{2x+3}}$$ Can the things in red cancel out? And would that lead me with a final answer?

6. Jamierox4ev3r

^This is where there are some clear holes in my recollection of simplifying. Someone clarify my thinking, thanks

7. jim_thompson5910

you are correct. The 2x+3 terms will divide to 1 and effectively go away since 1*(any number) = same number the same applies to the x+5 terms

8. Jamierox4ev3r

so $$\huge\frac{5x+7}{x-5}$$ would be final answer? I mean, I'm not sure if that could be simplified any further, but I don't think so.

9. jim_thompson5910

that's as simplified as it gets

10. Jamierox4ev3r

Hmm alright. Thank you for your time :)

11. jim_thompson5910

you're welcome