# I have a small problem with long division when dividing polynomials of the same degree

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.

• 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. Jan 18, 2022 at 20:51
• Then the quotient is $1/3$, no? Jan 18, 2022 at 21:00
• 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." Jan 18, 2022 at 21:05
• Terminology: $\dfrac{\text{dividend}}{\text{divisor}} = \text{quotient}$ Jan 18, 2022 at 21:06
• "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 $3$. Jan 18, 2022 at 21:38

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