Question

Divide 6x^4 + 5x^3 + x^2 - 3x + 1 by 3x + 1?

Answer 1

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3x + 1 | 6x^4 + 5x^3 + x^2 - 3x + 1

You work these just like long division on numbers, except it turns out to be easier!

What is \(6x^4/3x\) (first term of numerator divided by first term of denominator)

That's the first term of your answer: \(2x^3\)

Now we multiply that term by the denominator: \(2x^3 *(3x+1) = 6x^4+2x^3\) and we subtract it from the numerator:

2x^3
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3x+1 | 6x^4 + 5x^3 + x^2 - 3x + 1
-6x^4 - 2x^3
---------------
3x^3 + x^2 - 3x + 1

repeat the process:
\(3x^3/3x = x^2\) so \(x^2\) is next term in quotient. Multiply \(x^2(3x+1) = 3x^3+x^2\) and subtract from remaining polynomial:

2x^3 + x^2
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3x+1 | 6x^4 + 5x^3 + x^2 - 3x + 1
-6x^4 - 2x^3
---------------
3x^3 + x^2 - 3x + 1
-3x^3 -x^2
-------------
-3x + 1

Here things get slightly tricky: our next term is -1 because \(-3x/3x = -1\)

2x^3 + x^2 -1
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3x+1 | 6x^4 + 5x^3 + x^2 - 3x + 1
-6x^4 - 2x^3
---------------
3x^3 + x^2 - 3x + 1
-3x^3 -x^2
-------------
-3x + 1
-(-3x - 1)
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2

it didn't come out evenly, so our remainder is 2, and our quotient is
\[2x^3+x^2-1 + \frac{2}{3x+1}\]

Answer 2

You may be thinking "why does he think this is easier?" Well, you never have to do that foolishness where you guess the next digit of the answer, multiply by the denominator, and discover that it's too large and have to pick a smaller one. And you don't have to worry about writing down the next digit in the proper column, or inserting 0s.