# 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 at 20:51
• Then the quotient is $1/3$, no? Jan 18 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 at 21:05
• Terminology: $\dfrac{\text{dividend}}{\text{divisor}} = \text{quotient}$ Jan 18 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 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 $$\frac {x^2+1}{3x^2 + 5} = \frac 13 -\frac {\frac 23}{3x^2 + 5}$$.