## anonymous one year ago MEDALS PLEASE HELP!! Please subtract and simplify 4y/y^2-6y+8 - 16/y^2-6y+8

1. anonymous

$\frac{ 4_{y} }{ y ^{2}-6_{y}+8 } - \frac{ 16 }{ y ^{2}-6_{y}+8 }$ This right?

2. anonymous

Yes

3. anonymous

Since the denominators are exactly the same, wouldn't the equation just be $\frac{ 4_{y} -16 }{ y ^{2}-6_{y} + 8}$

4. anonymous

It told me it has to be simplified down

5. anonymous

rheaanderson341 is correct ^

6. anonymous

So the correct answer is 4y-16/y^2-6y+8 ???

7. anonymous

No, it still can be simplified. You can simplify the $4_{y} and 6_{y}$ along with the 16 and 8.

8. anonymous

$4\div(y-2) simplified$

9. anonymous

|dw:1442451532151:dw|

10. anonymous

(y-4) cancels out on top and bottom

11. anonymous

So its 4/y-2 right?

12. anonymous

yeah

13. anonymous

Wow! I forgot about that method. What's it called again, @mshaw24 ?

14. anonymous

I understand that now! Thank you so much! Do you by any chance know anythinf about finding the real zeros of a function

15. anonymous

Its just expanding out the polynomial on the bottom and then cancelling it out

16. anonymous

Possibly its been a while since ive done it. whats the question

17. anonymous

F(x)= x^3 +4x^2+x-6

18. anonymous

It says there is 3 real zeros

19. anonymous

so when finding zeros, you have the amount as the highest power. IE x^3 = 3 real zeros as to finding them set the equation = to zero and solve.

20. anonymous

$x(x ^{2}+4x+1) = 6$

21. anonymous

zeros are x=6 and then expand out the polynomial

22. anonymous

hang on solving it now

23. anonymous

Okay thank you!

24. anonymous

Have you learned quadratic formula?

25. anonymous

Yes

26. anonymous

$\frac{ -B+-\sqrt{B ^{2}-4AC} }{ 2A }$

27. anonymous

use this and it gives you -0.2679 and -3.732 plus 6 from above. i think these are your answers

28. anonymous

Would i have to round those answers because it keeps telling me its incorrect?

29. anonymous

try $4+\sqrt{3}$ and [4-\sqrt{3}\]

30. anonymous

if its not that then I am not sure. It might have been a simple mistake, but like i said its been a while

31. anonymous

Its still telling me no, ill have to research how to do it, thank for all of your help. Do you mind helping me with one final question?

32. anonymous

I actually have to get to bed. Sorry i have early classes haha best of luck to you