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Mathematics 11 Online
OpenStudy (anonymous):

Using synthetic division, find (3x2 + 6x − 4) ÷ (x + 1). 3x+3 3x2+3x−7 3x+3−7x+1 3(x+7)−3x+1

OpenStudy (anonymous):

3x+3+(7/(x+1))

OpenStudy (anonymous):

there's no answer like tht

OpenStudy (anonymous):

uhh yeah i see thinking

OpenStudy (anonymous):

@satellite73 , @StudyGurl14

OpenStudy (aum):

(3x2 + 6x − 4) / (x + 1) Synthetic division: -1 3 6 -4 -3 -3 --------- 3 3 -7 3x + 3 - 7/(x+1)

OpenStudy (anonymous):

i don't have an answer like tht

OpenStudy (studygurl14):

@aum 's answer is this: \(\large 3x + 3-\frac{7}{x+1}\) Are you sure you don't have something similar to this?

OpenStudy (aum):

\[ 3x + 3 - \frac{7}{x+1} \]

OpenStudy (studygurl14):

You might have something like R -7 instead of the fraction

OpenStudy (anonymous):

OMG yass, plz can you help me, i got more!

OpenStudy (studygurl14):

lol, i thought so. I got these type of problems before. :)

OpenStudy (aum):

I think it is the third choice but the poster has left out a "/". I think the third choice should be: 3x+3−7/(x+1)

OpenStudy (anonymous):

One of the factors of the polynomial x3 − 5x2 is x + 3. What is the other factor? x2−8x+24+72x+3 x2−8x+24 x2−8x+24−72x+3 x3−8x2+24x−72

OpenStudy (studygurl14):

Anyway, if you need more help please ask @aum or someone else. Unfortunately, I have to go eat dinner. Good luck @stellawalkerson :)

OpenStudy (anonymous):

@aum can you plz help

OpenStudy (aum):

One of the factors of the polynomial x^3 − 5x^2 is x + 3. What is the other factor? Divide (x^3 − 5x^2) by (x + 3). Do you know synthetic division? If you learn it you can do all these problems on your own. Here is one good tutorial: http://www.youtube.com/watch?v=bZoMz1Cy1T4 Divide (x^3 − 5x^2) by (x + 3) or Divide (1x^3 - 5x^2 + 0x + 0) by (x+3)

OpenStudy (anonymous):

Using synthetic division, find (2x4 − 5x3 + 7x + 9) ÷ (x + 2). 2x3−9x2+18x−29 2x3−9x2+18x−29+67x+2 2x3−9x2+18x−29−67x+2 2x4−9x3+18x2−29x+67

OpenStudy (aum):

Like I mentioned earlier, learn to do synthetic division just once and you will be able to answer all these problems on your own.

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