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Division with Polynomials

Division with polynomials looks complicated.  But you can do long division with polynomials in much the same way that you do division with numbers!  And, like with long division, you just have to pay attention to where you put your answer terms so that you can keep track of where you are in the problem. 

To do polynomial division, you use the long division "house."  Let's review that process very quickly:

The problem: $56 \div 4=$ is written: $4\overline{)56}$

It is often difficult for students to get used to the fact that $4\overline{)56}$ is said "56 divided by 4" -- but they should practice because part of understanding division is being able to say the problems correctly (and understanding that when a number is divided by another number, it's cut into that many pieces).

With the American "house" system we follow a common procedure in long division:

  1. Division: Divide the divisor into the first digit(s) of the dividend (write the answer on top of the house).
  2. Multiply: Multiply the answer number times the divisor (write the answer under the portion of the dividend you are dividing into).
  3. Subtract: Subtract the production of the last step from the portion of the dividend you are working with (write the answer below).
  4. Check: Make sure that answer to your subtraction problem is not greater than your divisor (if it is, then the number you wrote on the top line is too small).
  5. Bring down: Bring the next digit in the dividend down and make it the ones digit of the number you just checked against the divisor.
  6. Repeat: How many times does the divisor go into the new number you just created (write the answer on the top of the house, above the number you just brought down).

Some students use an acronym to remember the steps.  You can try one of the ones below or make up your own.  It's often helpful for beginning dividers to write the acronym on the side of their papers and check off the steps as they go along.

Bring downBatheBurgers

So, how do the steps work?

Example: $786\div 5=$

Set the problem up:

$$\eqalign{5&\overline{)786}\qquad&&\text{Write the problem in a house}}$$

Do the first "division" step.  How many times does 5 go into the first digit in the dividend?

$$\eqalign{ \quad & \; \color{red}{1}\qquad && \color{red}{\text{5 goes into 7 one time, write 1 over the 7}}\\5 & \overline{)786}\qquad && \quad \\&\!\! \underline{-5}  \qquad && \text{5 times 1 is five; subtract from 7}\\  &\;\; 2  \qquad && \text{7 minus 5 equals 2}}$$

Now, you bring down and divide into the new number created when you bring a new digit from the dividend and place it next to the remainder left over from the first time you multiplied and subtracted!

$$\eqalign{ \quad & \; \color{red}{1}\color{green}5\qquad && \color{green}{\text{5 goes into 28 five times, write the 5 over the 8}}\\5 & \overline{)786}\qquad && \quad \\&\!\! \underline{-5} \color{green}{\downarrow} \qquad && \\  &\;\; 2\color{green}{8}  \qquad && \text{Bring down the 8 and put it next to the 2}\\&\!\!\underline{-25}\qquad &&\text{5 times 5 is 25}\\&\;\;\;3\qquad&&\text{28 minus 25 is 3}}$$

There is one more digit in the dividend, so you'll bring down one more time, then divide again into the new number.

$$\eqalign{ \quad & \; \color{red}{1}\color{green}5\color{blue}{7}\qquad && \color{blue}{\text{5 goes into 36 seven times, write the 7 over the 5}}\\5 & \overline{)786}\qquad && \quad \\&\!\! \underline{-5} \color{green}{\downarrow} \qquad && \\  &\; 2\color{green}{8}  \qquad && \\&\!\!\!\underline{-25}\color{blue}{\downarrow}\qquad &&\\&\;\;3\color{blue}{6}\qquad&&\text{Bring down the 6}\\&\!\underline{-35}\qquad&&\text{7 times 5 is 35}\\&\quad1\qquad&&\text{36 minus 35 is 1, you have a remainder of 1}}$$

Bring the remainder up to the top.  Most elementary students use a small "r" to denote the remainder.  It's also helpful to show students that remainder can easily be turned into a fraction: $\dfrac{\text{remainder}}{\text{divisor}}$

The answer is now at the top of the "house."  $\color{purple}{785 \div 5 = 157 \text{ r }1}$ or $\color{purple}{785 \div 5 = 157 \dfrac{1}{5}}$

Looking at the problem above, you can see that you make your way through the dividend digit by digit, placing the answer digit ABOVE the last digit of the number you just into, in order to keep your place in the problem.  With polynomial division you will do the same exact thing, except you will divide term by term, placing your answer term above the last term of the binomial you just divided into in order to keep your place. 

Example: $x^2+17x-9\div x+2=$

Set the problem up:

$$\eqalign{x+2&\overline{)x^2+17x-9}\qquad&&\text{Write the problem in a house}}$$

Do the first "division" step.  Focus on the first term in your divisor (in this case $x$). How many times does $x$ go into $x^2$?  The other way to think about it is, what do you have to multiply $x$ by to make it $x^2$?

$$\eqalign{ \quad & \quad \qquad \color{red}{x}\qquad && \color{red}{\text{multiply }x \times x+2 \text{ to make }x^2 + 2x}\\x+2 & \overline{)x^2+17x-9}\qquad && \quad \\&\!\! \!\!\!\underline{-(x^2+\:\;2x)}  \qquad && x \times x+2 = x^2+2x\text{; subtract from }x^2+17x\\  &\qquad \;\; 15x  \qquad && x^2-x^2=0 \text{ and }17x-2x=15x}$$

Above, you should have noticed, you that you take your answer term and multiply it by BOTH terms in the binomial.  So, while the first term will zero out when you subtract, the second term ($17x-2x$) will leave a remainder.  Now, you bring down the next term in the dividend and you have a new binomial to divide into. Again, when you figure out how many times the divisor goes in, just focus on the first term.  What times $x$ will equal $15x$?

$$\eqalign{ \quad & \quad \qquad\color{red}{x}\qquad && \\x+2 & \overline{)x^2+17x-9}\qquad && \quad \\&\!\! \!\!\!\underline{-(x^2+\:\;2x)} \;\color{green}{\downarrow} \\  &\qquad\;\; 15x  \; \color{green}{-9}  \qquad && \color{green}{\text{Bring down the -9 and put it next to the 15x}}}$$

Now, how many times does $x+2$ go into your new number ($15x-9$)? Put your answer in the answer row, above the $-9$, which is the last term you just brought down. Then multiply the 15 times $x+2$ adn subtract it from $15x-9$.

$$\eqalign{ \quad & \quad \qquad\color{red}{x}\;\color{green}{+15}\qquad && \\x+2 & \overline{)x^2+17x-9}\qquad && \quad \\&\!\! \!\!\!\underline{-(x^2+\:\;2x)} \;\color{green}{\downarrow} \\  &\qquad\;\; 15x  \; \color{green}{-9}\\&\qquad\!\! \underline{-(15x  \; +30)}&&\text{Multiply } 15 \times x+2=15x+30\\  &\qquad \qquad \quad \color{blue}-39 &&\color{blue}{\text{Subtract }15x+30 \text{ from }15x-9=-39}}$$

There is a -39 left over.  That's your remainder.  Just as you did in the long division problem below, you can put that remainder over the divisor to make a fraction.  The final answer:

$\large{x^2+17x-9 \div x+2 = \color{blue}{x+15- \dfrac{39}{x+2}}}$

Polynomial division takes practice.  But, the process is exactly the same as long division, so keep working at it and it will become sort of fun!

Practice Problems:

  • Division with Polynomials

    1. $\dfrac{x^2+12x+27}{x+3}=$
    2. $\dfrac{x^2-7x+10}{x-2}=$
    3. $\dfrac{2x^2+3x-1}{x+2}=$
    4. $\dfrac{5x^2-5x+3}{x+3}=$
    5. $\dfrac{x^2+17x-9}{x+2}=$
    6. $\dfrac{x^2+21x-3}{x+5}=$
    7. $\dfrac{3x^2+11x+4}{x+3}=$
    8. $\dfrac{x^2+x+1}{x+9}=$
    9. $\dfrac{2x^2+4x-8}{x+3}=$
    10. $\dfrac{x^2+7x+2}{x+1}=$
    11. $\dfrac{x^2-5x-6}{x-1}=$
    12. $\dfrac{x^2+17x-9}{x+2}=$
    13. $\dfrac{3x^2+4x-5}{x+1}=$
    14. $\dfrac{12x^2-40x+63}{x+4}=$


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