I was doing long division with $x^2 + 1$, and $3x^2+5$. (the second polynomial is the quotient). the problem I've found is related to dividing two polynomials of the same degree. Even if I know that the quotient is always a constant and the remainder is a polynomial of one degree less than the dividend, I still have problem when performing this long division, meaning:

$x^2+0+1\space /\space 3x^2+5$

x^2 is contained in 3x^2 3 times, so I write 3 in the quotient.

then, I perform multiplication between the quotient and the divisor. $3 * 3x^2 = 9x^2$, and $3*5 = 15$. I write them below the dividend. Now, I subtract the dividend with the things I have below.

but, it's an infinite loop, because the degree doesn't change no matter how long I divide.

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    $\begingroup$ When you multiply the quotient and the divisor, isn't the divisor $x^2 + \cdots$ rather than $3x^2 + 5$? Multiplying $x^2 + \cdots$ by $3$ yields $3x^2 + \cdots$ which will cancel with the leading term of $3x^2+5$ when subtracting. $\endgroup$
    – angryavian
    Jan 18 at 20:51
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    $\begingroup$ Then the quotient is $1/3$, no? $\endgroup$
    – angryavian
    Jan 18 at 21:00
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    $\begingroup$ In general you usually ask "how many times is the divisor contained in the dividend," rather than "how many times is the dividend contained in the divisor." $\endgroup$
    – angryavian
    Jan 18 at 21:05
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    $\begingroup$ Terminology: $\dfrac{\text{dividend}}{\text{divisor}} = \text{quotient}$ $\endgroup$ Jan 18 at 21:06
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    $\begingroup$ "because as I wrote above x^2 is contained in 3x^2 3 times; am I wrong?" You are right if you are dividing $3x^2 + 5$ but $x^2 +1$ which is not what you wrote. If you wrote $x^2 + 1$ divided by $3x^2 + 5$ then the quotient is $\frac 13$ because $3x^2$ is contained in $x^2$ $\frac 13$ times. If you are dividing by $x^2 + 1$ then you are supposed to multiply $x^2 +1$ by $3$; not $3x^2 + 5$ by $3$. $\endgroup$
    – fleablood
    Jan 18 at 21:38

1 Answer 1


You are confusing what you dividing into with what you are dividing by and you are taking the quotient, $3$, and multiply it by what you are dividing into; not what you are dividing by. You must multiply the quotient by what you are dividing by.

Question 1: $\frac {3x^2+ 5}{x^2 + 1}$ then we divide $x^2$ into $3x^2$ and get a quotient of $3$. So we multiply the denominator, $x^2 + 1$ by $3$ to get $3(x^2+1)=3x^2 + 3$. Then you subtract $(3x^2 + 5)-(3x^2 + 3) = 2$. Now you have the remainder.

So $\frac {3x^2 + 5}{x^2 + 1} = 3 + \frac 2{x^2 + 1}$.

Question 2: $\frac {x^2 + 1}{3x^2 + 5}$ then we divide $3x^2$ into $x^2$ and we get a quotient of $\frac 13$ (because $3x^2$ goes into $x^2$ a total of $\frac 13$ times). So we multiply the denominator, $3x^3 + 5$ by $\frac 13$ to got $\frac 13(3x^2 + 5) = x^2 + \frac 53$. Then we subtract $(x^2 + 1)-(x^2 + \frac 53)= -\frac 23$. Now we have a remainedr of $-\frac 23$.

So $\frac {x^2+1}{3x^2 + 5} = \frac 13 -\frac {\frac 23}{3x^2 + 5}$.


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