Answer: 2
Step by step: I start with a long-division set-up
Looking only at the leading terms, I divide 3x3 by 3x to get x2. This is what I put on top:
x^2 up on top
I multiply this x2 by the 3x + 1 to get 3x3 + 1x2, which I put underneath:
3x^3 + 3x put down underneath the dividend
Then I change the signs, add down, and remember to carry down the "+10x – 3" from the original dividend, giving me a new bottom line of –6x2 + 10x – 3:
(3x^3 – 5x^2 + 10x – 3) – (3x^3 + 1x^2) = –6x^2 + 10x – 3
Dividing the new leading term, –6x2, by the divisor's leading term, 3x, I get –2x, so I put this on top:
(–6x^2) ÷ (3x) = –2x, which goes on top
Then I multiply –2x by 3x + 1 to get –6x2 – 2x, which I put underneath. I change signs, add down, and remember to carry down the "–3 from the dividend:
(–6x^2 + 10x –3) – (–6x^2 – 2x) = 12x – 3, which is my new last line
My new last line is "12x – 3. Dividing the new leading term of 12x by the divisor's leading term of 3x, I get +4, which I put on top. I multiply 4 by 3x + 1 to get 12x + 4. I switch signs and add down. I end up with a remainder of –7:
(12x)/(3x) = 4, 4(3x + 1) = 12x + 4, (12x – 3) – (12x + 4) = –7
This division did not come out even. What am I supposed to do with the remainder?
Think back to when you did long division with plain numbers. Sometimes there would be a remainder; for instance, if you divide 132 by 5:
132 ÷ 5: 2 × 5 = 10; put 2 on top and 10 underneath; 13 – 10 = 3; new divisor is 32; 6 × 5 = 30; put 6 on top and 30 underneath; 32 – 30 = 2; answer is 26 with remainder 2
...there is a remainder of 2. Remember how you handled that? You made a fraction, putting the remainder on top of the divisor, and wrote the answer as "twenty-six and two-fifths":
\dfrac{132}{5} = 26\,\dfrac{2}{5} = 26 + \dfrac{2}{5}
5
132
=26
5
2
=26+
5
2
The first form, without the "plus" in the middle, is how "mixed numbers" are written, but the meaning of the mixed number is actually the form with the addition.
We do the same thing with polynomial division. Since the remainder in this case is –7 and since the divisor is 3x + 1, then I'll turn the remainder into a fraction (the remainder divided by the original divisor), and add this fraction to the polynomial across the top of the division symbol. Then my answer is this:
\mathbf{\color{purple}{\mathit{x}^2 - 2\mathit{x} + 4 + \dfrac{-7}{3\mathit{x} + 1}}}x
2
−2x+4+
3x+1
−7
