## Jamierox4ev3r one year ago Simplify completely:

1. anonymous

hi again

2. Jamierox4ev3r

$$\Large 9(\frac{t-2}{2t+1})^{8}\times \frac{(2t-1)-2(t-2)}{(2t+1)^{2}}$$

3. anonymous

i think u know me as GTA_Hunter35 that account kinda crashed

4. Jamierox4ev3r

^^There's the thing I have to simplify. Not quite sure where to begin.

5. Jamierox4ev3r

6. anonymous

Parentheses Exponents Multiplication Division Addition Subtraction FROM LEFT TO RIGHT

7. Michele_Laino

all we can do is: $\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{3}{{{{(2t + 1)}^2}}} \hfill \\ \end{gathered}$

8. Jamierox4ev3r

Is that all we can do?

9. Michele_Laino

therefore, we can write this: $\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{3}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 27\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\ \end{gathered}$

10. Michele_Laino

yes! I think so!

11. anonymous

12. Michele_Laino

better is if we don't develop the powers of binomials, because we have to do a very long computation @please_help_me

13. anonymous

well.... ur right ithink thats it

14. Jamierox4ev3r

wait. For the top part, (2t+1)-2(t-2) = 2t-2t+1+4? so then we would have $$\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{\color{red}{5}}{{{{(2t + 1)}^2}}} \hfill \\ \end{gathered}$$

15. Jamierox4ev3r

basically, a 5 instead of a 3?

16. Michele_Laino

I got 2t-1-2t+4

17. anonymous

which equals 3

18. anonymous

distribute, then ull get 3

19. Jamierox4ev3r

-1? o-o oh. wait, i think i see something. $$\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t \color{red}+ 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{\color{red}{5}}{{{{(2t + 1)}^2}}} \hfill \\ \end{gathered}$$

20. Jamierox4ev3r

in the original problem, it is 2t+1, not 2t-1.

21. Jamierox4ev3r

oh wow. I wrote it out wrong. So sorry XD

22. anonymous

np was a typo

23. Michele_Laino

ok! then you are right, it is 5

24. Michele_Laino

here are the updated steps: $\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t + 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{5}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\ \end{gathered}$

25. Jamierox4ev3r

to clarify, let me type this out correctly. *cracks knuckles* $$\LaTeX$$ is hard $$\Large 9(\frac{t-2}{2t+1})^{8} \times \frac{(2t+1)-2(t-2)}{(2t+1)^{2}}$$

26. anonymous

can u guys help me

27. Jamierox4ev3r

So I don't have to factor out the denominator or do anything fancy like that?

28. anonymous

i dont think so

29. Michele_Laino

no, since numerator and denominator don't contain common divisors

30. Jamierox4ev3r

oh, and how did you get from the penultimate step to the last step? Like what are the exact operations that are involved in combining this multiplication problem into one fraction?

31. Jamierox4ev3r

and yeah you're right, I don't see any common divisors. That makes sense

32. anonymous

see the 2t-1?

33. Michele_Laino

I have applied the rule of multiplication of powers with same basis at denominator

34. anonymous

2t-1 was to the 8th power and the other 2t-1 was to the2nd power

35. anonymous

multiplying numbers with exponents adds up the exponents

36. Michele_Laino

here are more steps: $\large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t + 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{5}{{{{(2t + 1)}^2}}} = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^8}}} \cdot \frac{1}{{{{\left( {2t + 1} \right)}^2}}} \hfill \\ \hfill \\ = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\ \end{gathered}$

37. anonymous

which created 2t-1^10

38. Jamierox4ev3r

Oh I see! I was just kind of wondering how the numerator of five contributed to the solid number of 45 in the mixed fraction. I see it now

39. Michele_Laino

ok! :)

40. Jamierox4ev3r

Thank you so much, it makes sense now :D

41. Michele_Laino

:)