SOLUTION: Use polynomial division to determine: 3x^3-5x^2+10x+4 / 3x+1 Any help would be appreciated I have not done these for years and cant get my head around them Thanks

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Use polynomial division to determine: 3x^3-5x^2+10x+4 / 3x+1 Any help would be appreciated I have not done these for years and cant get my head around them Thanks      Log On


   



Question 699219: Use polynomial division to determine:
3x^3-5x^2+10x+4 / 3x+1
Any help would be appreciated I have not done these for years and cant get my head around them
Thanks

Found 2 solutions by stanbon, pmatei:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Use polynomial division to determine:
3x^3-5x^2+10x+4 / 3x+1
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Divide the dividend by the divisor:
1st term of the quotient: x^2
----
Multiply times the divisor to get:
3x^3+x^2
----
Subtract from the dividend to get:
-6x^2+10x+4
----
Divide that dividend by the divisor:
2nd term of the quotient: -2x
===============
Multiply times the divisor to get:
-6x^2-2x
---
Subtract from the dividend to get:
12x+4
---
Divide the dividend by the divisor to get
3rd term of the quotient: 4
====
Multiply times the divisor to get:
12x + 4
-----
Subtract from the dividend to get:
0
=====
Final Quotient: x^2 -2x + 4
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Cheers,
Stan H.
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Answer by pmatei(79) About Me  (Show Source):
You can put this solution on YOUR website!
So your Numerator is N=3x%5E3-5x%5E2%2B10x%2B4 and your Denominator is D=3x%2B1.
I will write this as a fraction, and explain how I get every term in the Quotient.
N%2FD=%283x%5E3-5x%5E2%2B10x%2B4%29%2F%283x%2B1%29
The first term of the Quotient is given by the division of the first term in Numerator 3x%5E3 by the first tern in the Denominator 3x, which is
Q1=%283x%5E3%29%2F%283x%29=x%5E2
Then I multiply the term I just found x%5E2 with the Denominator and the result is subtracted from the Numerator to calculate the Remainder:
Q1%2AD=x%5E2%283x%2B1%29+=+3x%5E3%2Bx%5E2
The Remainder is:
R1=N-%28Q1%2AD%29=%283x%5E3-5x%5E2%2B10x%2B4%29-%283x%5E3%2Bx%5E2%29=-6x%5E2%2B10x%2B4
So I can write N%2FD+=+Q1+%2B+R1%2FD:
%283x%5E3-5x%5E2%2B10x%2B4%29%2F%283x%2B1%29=x%5E2%2B%28-6x%5E2%2B10x%2B4%29%2F%283x%2B1%29
Now I do it again for the fraction %28-6x%5E2%2B10x%2B4%29%2F%283x%2B1%29
Q2=%28-6x%5E2%29%2F%283x%29=-2x
Q2%2AD=-2x%283x%2B1%29+=+-6x%5E2-2x
R2=%28-6x%5E2%2B10x%2B4%29-%28-6x%5E2-2x%29=12x%2B4
So now my quotient is Q=Q1%2BQ2=x%5E2-2x
%283x%5E3-5x%5E2%2B10x%2B4%29%2F%283x%2B1%29=x%5E2-2x%2B%2812x%2B4%29%2F%283x%2B1%29
Do it one more time for fraction %2812x%2B4%29%2F%283x%2B1%29
Q3=%2812x%29%2F%283x%29=4
Q3%2AD=4%283x%2B1%29=12x%2B4
R3=%2812x%2B4%29-%2812x%2B4%29=0
Q=Q1%2BQ2%2BQ3=x%5E2-2x%2B4
So the final result is:
%283x%5E3-5x%5E2%2B10x%2B4%29%2F%283x%2B1%29=x%5E2-2x%2B4