MEDALS PLEASE HELP!!
Please subtract and simplify
4y/y^2-6y+8 - 16/y^2-6y+8

- anonymous

MEDALS PLEASE HELP!!
Please subtract and simplify
4y/y^2-6y+8 - 16/y^2-6y+8

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- anonymous

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

- anonymous

Yes

- anonymous

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

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## More answers

- anonymous

It told me it has to be simplified down

- anonymous

rheaanderson341 is correct ^

- anonymous

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

- anonymous

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

- anonymous

\[4\div(y-2) simplified\]

- anonymous

|dw:1442451532151:dw|

- anonymous

(y-4) cancels out on top and bottom

- anonymous

So its 4/y-2 right?

- anonymous

yeah

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

It says there is 3 real zeros

- 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.

- anonymous

\[x(x ^{2}+4x+1) = 6\]

- anonymous

zeros are x=6 and then expand out the polynomial

- anonymous

hang on solving it now

- anonymous

Okay thank you!

- anonymous

Have you learned quadratic formula?

- anonymous

Yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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

- 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?

- anonymous

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

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