changing to mixed expression

[bronx.system]

i kept trying to mess around with this and i cant match the book answer. Any chance someone can confirm if book has right answer?

change to mixed expression

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

i keep getting answer 2x+8+ 15/(x+2)

not sure how to get latex to work here so i took screen shot. The bottom one is my answer

 

[wjm11]

i kept trying to mess around with this and i cant match the book answer. Any chance someone can confirm if book has right answer?

change to mixed expression

[math] \frac{2x^2-4x+1}{x+2}[/math]
i keep getting answer 2x+8+ 15/x+2

not sure how to get latex to work here so i took screen shot. The bottom one is my answer

You can easily check your answer yourself. Just multiply each term in your answer by (x+2) and see if you get the numerator of your original problem.

(x + 2)(2x + 8 + 15/(x+2)) = 2x^2 + 4x + 8x + 16 + 15 = 2x^2 + 12x + 31.

No, the check does not work; your solution is incorrect.

Your problem lies in some misinterpreted/mixed up signs. Check the subtractions in your long division. For example, your first step:

(2x^2 - 4x) - [2x(x + 2)] = (2x^2 - 4x) - [2x^2 +4x] = -8x

I'll bet you got +8x.

Hope that helps.

 

[mmm4444bot]

i keep getting answer 2x+8+ 15/(x+2)

We follow the Order of Operations, at this site.

not sure how to get latex to work here

Use tex and /tex, instead of math and /math.


Note: \dfrac{}{} displays better than \frac{}{}


Cheers :cool:

 

[bronx.system]

hey all thanks sorry i didnt check myself it was knee jerk reaction i just couldnt match the books answer so ran to forum in frustration >< i will check answers myself in future sorry again.

i messed up the signs /= the signs made me slip up i need write it out my neatly and have more practice

thanks all very much appreciated

 

[lookagain]

Check YOURSELF if correct; easiest way is let x=1:
top: (2 - 4 + 1) / 3 = -1/3

yours: 2 + 8 + 15/3 = 15

So you're way off...

Denis, if one particular x-value had checked, that would not have proven that that was the equivalent correct mixed expression \(\displaystyle \ \ \ \) <--- edit
for the given expression.