Long Division to Divide Polynomials Practice

Note that $b(x)$ has a higher degree than $a(x)$. This implies that the quotient term, $q(x)$ is $0$. 

We already know that $q(x)=0$. Since the degree of $a(x)$ is smaller than the degree of $b(x)$, the remainder $r(x)$ is simply $a(x)$. 

Therefore, $r(x)=5x^3-6x^2-8x+9$.

To conclude,

• $q(x)=0$
• $r(x)=5x^3-6x^2-8x+9$

Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. 

Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of 

$\ \dfrac{a(x)}{b(x)}=\dfrac{-5x^3-x^2+3}{x^2+4}$:

First, we divide ${x^2}$ into ${-5x^3}$ and get ${-5x}$:

$\hphantom{1567|14} {-5x}\\{{{x^2}+4}}|\overline{{-5x^3}-x^2\ +\ 0x+\ 3}\\\hphantom{37{.}{.}{.}{.}|}\llap­\underline{(-5x^3+0x^2-20x)}\\\hphantom{37|3{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}}-x^2+\ 20x\\$

Next, we divide ${x^2}$ into ${-x^2}$ to get ${-1}$:

$\hphantom{1567|14} {-5x \ {- \ 1}}\\{{{x^2}+4}}|\overline{-5x^3-x^2\ +\ 0x\ +3}\\\hphantom{37{.}{.}{.}{.}|}\llap­\underline{(-5x^3+0x^2-20x)}\\\hphantom{37|3{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}}{-x^2}+\ 20x+3\\\hphantom{37{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}|}\llap­\underline{(-x^2\ +\ 0x\ - 4)}\\\hphantom{37|3{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}} {+20x\ +\ 7}\\$

The process stops here because $x^2+4$ is a polynomial of the second degree and $20x+7$ is a polynomial of the first degree. So it follows that
${r(x)}= {20x+7}$, ${q(x)}= {-5x-1}$, and

$\qquad\dfrac{-5x^3-x^2+3}{x^2+4}= {-5x-1}+\dfrac{ {20x+7}}{x^2+4}$

To conclude,

• $q(x)=-5x-1$
• $r(x)=20x+7$

Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. 

Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of 

$\ \dfrac{a(x)}{b(x)}=\dfrac{-12x^5-2x^3-9x}{3x^4+x^2+1}$:

We divide ${3x^4}$ into ${-12x^5}$ to get ${-4x}$:

$\hphantom{1567|1444477} { {-4x}}\\{{{3x^4}+x^2+1}}|\overline{{-12x^5}-2x^3-9x}\\\hphantom{37{.}{.}{.}88{.}{.}{.}{.}{.}{.}{.}{.}|}\llap­\underline{(-12x^5-4x^3-4x)}\\\hphantom{37|3{.}{.}{.}{.}99888889{.}{.}{.}{.}{.}{.}} {+2x^3-5x }\\$

The process stops here because $3x^4+x^2+1$ is a polynomial of the fourth degree and $2x^3-5x$ is a polynomial of the third degree. So it follows that ${r(x)}= {2x^3-5x}$, ${q(x)}= {-4x}$, and

$\qquad\dfrac{-12x^5-2x^3-9x}{3x^4+x^2+1}= {-4x}+\dfrac{ {2x^3-5x}}{3x^4+x^2+1}$

To conclude,

• $q(x)=-4x$
• $r(x)=2x^3-5x$

Note that $b(x)$ has a higher degree than $a(x)$. This implies that the quotient term, $q(x)$ is $0$. 

We already know that $q(x)=0$. Since the degree of $a(x)$ is smaller than the degree of $b(x)$, the remainder $r(x)$ is simply $a(x)$. 

Therefore, $r(x)=5x^2-6x+10x-2$.

To conclude,

• $q(x)=0$
• $r(x)=5x^2-6x+10x-2$