Polynomial Long Division Calculator

Polynomial long division is a method/technique by which we can divide a polynomial by another polynomial of the same or a lower degree.
Division of a polynomial `(ax^2 + bx + c) ` by another polynomial (dx + e) can be expressed in the form:

`(ax^2 + bx + c) / (dx + e)`

Where a,b,c,d and e are any constant values.

The polynomial on the top is called the "numerator" whereas the polynomial on the bottom is termed as "denominator". These terms are useful to remember, as we will use them frequently in the coming text. (Note: Remember denominator from down).

While dividing using Long Division method, we write the numerator and the denominator like this:


Polynomial Long Division

Steps for Division

1. Divide the leading term of the numerator ( `x^2` in our case ) with the leading term of the numerator ( x in this example. ) This gives us a value equal to x. We place this value on top of the horizontal bar as shown below:

_________
` x + 2 ) x^2 - 3x - 10`


2. The next step is to multiply the x on top across the divisor. We first multiply x with x, this results in `x^2`. We place this `x^2` underneath the `x^2` in the numerator.

_________
`x + 2 ) x^2 - 3x - 10 `


3. Next we will multiply the x on top of the horizontal bar with +2 in the divisor. This produces +2x. We place this term right underneath - 3x inside the division.

_________
`x + 2 ) x^2 - 3x - 10 `
`x^2 + 2x `


4. Now the next step is to subtract the lower terms inside the division from the terms above. Remember we learned in the previous sections that subtraction requires inversion of all the signs in the lower polynomial. We will do the same here.

`x`
_________
` x + 2 ) x^2 - 3x - 10`
`- x^2 - 2x`

After we have inverted all the signs in the lower polynomial, we simply need to add the like terms as shown in the step below:

`x`
_________
`x + 2 ) x^2 - 3x - 10`
`- x^2 - 2x`
_____________
`- 5x - 10``


5. As shown in the step above, the ` x^2` terms have been cancelled out. - 3x - 2x add together to give - 5x. And then we carry the last term in the numerator down as it is. The next step is to divide (- 5x - 10) by (x + 2)


6. Just like we did above, we divide the leading term in the dividend (- 5x) by the leading term in the divisor (which is the same as before: x) This division produces - 5. We place this - 5 on top of the horizontal bar next to x.

` x - 5`
________
`x + 2 ) x^2 - 3x - 10`
`- x^2 - 2x`
_____________
`- 5x - 10`


7. The next step is to multiply each term in the divisor by this -5 and to place the like terms obtained underneath the dividend terms as shown below:

` x - 5`
________
`x + 2 ) x^2 - 3x - 10`
`- x2 - 2x`
_____________
`- 5x - 10`
`- 5x - 10`


8. Next we subtract the lower terms from the terms above. Inversion of signs in the bottom row produces:

Polynomial Long Division


9. We saw that the two terms got cancelled. Hence we got remainder = 0 What we are interested in is the quotient and the remainder. For our example:

Quotient: x - 5

Remainder: 0


Note: We need to emphasize here that since the remainder of this division has come out to be zero, this shows that both the divisor (x + 2) and the quotient (x - 5) are the factors of the given dividend (x2 - 3x - 10).

Another interesting point to mention is that you can also verify your answer by multiplying the factors back. If you reach the dividend polynomial again, this means that you have done the right calculations

Let us show this by using our well-known FOIL method:

`(x - 5)(x + 2)`

= `x^2 + 2x - 5x - 10`

= `x^2 - 3x - 10 `

This verifies our answer.


Will the Remainder Always come out to be Zero? What happens if it doesn't?

This is not necessary for the remainder to always come out to be equal to zero. There are cases where we reach non-zero remainders. Such a situation arises when the divisor is not a factor of the dividend polynomial.

If in a polynomial division, we have divisor = Q, dividend = P, quotient = q, and remainder = r

Then, we can represent our answer as follows:

` ( P )/Q = q + ( r )/Q`

This implies that our original dividend polynomial can be expressed as:

` P = qQ + r - - - eq 1`

which is quotient times divisor plus the remainder.


Polynomial Division Examples


Example 1:

Explanation:

We solve this question using long division method.

_________
`x - 2 ) 2x^2 - 4x + 1 `

We divide the leading term of the numerator by the leading term in the denominator:

`2x^2 / x = 2x`

We multiply 2x across the denominator.

` 2x`
_________
`x - 2 ) 2x^2 - 4x + 1 `
- `2x^2 + 4x`
________________
+`1 `

Hence the remainder = 1 i.e., it is nonzero. If such a situation arises, we say that the denominator is not a factor of the numerator polynomial.

You can make use of eq 1 to verify your answer.

  • `P = qQ + r`

    `= (2x)(x - 2) + 1 `
  • ` = 2x^2 - 4x + 1 `

Which is the given numerator polynomial. It shows that our answer is correct.

We can write the answer of the polynomial division as follows:

`2x^2 - 4x + 1 ` = `2x`(` x - 2`)+ `1`

The process of polynomial division becomes very easy when the numerator involves only one term. In this case, we simply divide each term of the numerator by this term to get the answer. We present an example of this case as follows:


Example 2:

Explanation:

`(3x + 6)/3` = `(3(x + 2))/3`= (`x + 2`)


Example 3:

Explanation:

We can use the same procedure as above to solve this problem. We separate the two terms in the numerator and divide each by the denominator.

` (x(x + 1) - 4(x + 1))/(x + 1)` = `((x + 1)(x-4))/(x+1)` =(` x - 4 `)

Hence x - 4 is the required result.

Till this point, we have only divided

1) A second degree polynomial by a first degree polynomial.

2) A first degree polynomial by a first degree polynomial.

But there is no limitation to it. In Algebra, we may have to divide

3) A 3rd degree polynomial by a 1st degree polynomial.

4) A 3rd degree polynomial by a 2nd degree polynomial.

5) A 4th degree polynomial by a second degree polynomial.

And so on. As we proceed to higher degrees, the division process becomes more and more complex. We present a few examples of this type of long division in the next section.


Example 4:

Explanation:

In this example, we are going to divide a 3rd degree polynomial by a 1st degree polynomial. The procedure remains the same. The only difference is that a greater number of steps are involved in the process of division.

___________
` 2x + 1 ) 2x^3 + 5x^2 - 4x + 2`

Dividing the leading term of the numerator (2x3) by the leading term of the denominator (2x) produces x2. We multiply x2 across the divider, and subtract the resulting polynomial from the numerator terms as we did in the previous examples:

` x^2`
___________
`2x + 1 ) 2x^3 + 5x^2 - 4x + 2`
`2x^3 + x^2`
-       -
________________
`4x^2 `

We also take the term - 4x down. We will use this term in the next step of division:

` x^2`
___________
`2x + 1 ) 2x^3 + 5x^2 - 4x + 2`
`2x^3 + x^2`
________________
`4x^2 - 4x`

The next step is to repeat all the steps done above. We divide the leading term of `4x^2 - 4x` by the leading term of the divider. And so on.

` x^2 + 2x`
___________
`2x + 1 ) 2x^3 + 5x^2 - 4x + 2`
`2x^3 + x^2`
________________
`4x^2 - 4x`
`4x^2 + 2x`
-       -
________________
`- 6x + 2`

We have simplified the terms and have also taken the +2 term down. Dividing - 6x by 2x gives - 3. We multiply - 3 across the divider and repeat the above steps to reach our final solution:

` x^2 + 2x-3`
___________
`2x + 1 ) 2x^3 + 5x^2 - 4x + 2`

` x^2 + 2x`
___________
`2x + 1 ) 2x^3 + 5x^2 - 4x + 2`
`2x^3 + x^2`
________________
`4x^2 - 4x`
`4x^2 + 2x`
-       -
________________
`- 6x + 2`
`- 6x - 3 `
`+        +`
________________
+ 5 `

We can write the solution to our division problem as follows:

` ( 2x^3 + 5x^2 - 4x + 2 ) =( x^2 + 2x - 3 )(2x + 1) +5)`


Point to Ponder


"Whenever we divide a polynomial of degree n by a polynomial of degree m, the resulting polynomial has a degree (n - m)"

 Numerator with Missing Terms:

Until this point, we have only considered numerator polynomials that have all the terms present in them. For example if the degree of the polynomial is 4, we have all the `x^4, x^3, x^2, x^1` and constant terms present. But this is not the case always. Sometimes a numerator polynomial may have some of the intermediate terms missing.


Example 5:

Explanation:

` x^2 + 2x ) x^6 + 0x^5 + 3x^4 + 0x^3 - x^2 + 0x + 2`

Note that we have written the missing terms with 0 coefficient. Next we start the process of division in the same way as we did in the previous examples.

Polynomial Long Division
Polynomial Long Division

Hence we can write the answer of our division as follows:

` x^6 + 3x^4 - x^2 + 2 =(x^4 - 2x^3 + 7x^2 - 14x + 27)(x^2 + 2x)` `- 54x + 2`

Let us verify our answer by using equation (1).

P = qQ + r

` = (x^4 - 2x^3 + 7x^2 - 14x + 27)(x^2 + 2x) + (- 54x + 2) `

` = x^2(x^4 - 2x^3 + 7x^2 - 14x + 27) + 2x(x^4 - 2x^3 + 7x^2 - 14x + 27) + (- 54x + 2)`

`= x^6 - 2x^5 + 7x^4 - 14x^3 + 27x^2 + 2x^5 - 4x^4 + 14x^3 - 28x^2 + 54x - 54x + 2`

Simplification gives

` = x^6 + 3x^4 - x^2 + 2`

Which is the same as the given polynomial. Hence our result is correct.






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