# Dividing Polynomials

While dividing a polynomial f(x) by a non zero polynomial g(x), there exist two unique polynomials Q(x) and R(x) such that f(x) = g(x).Q(x) + R(x).

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Here, f(x) is dividend, g(x) is divisor, Q(x) is quotient and R(x) is remainder.

If R(x) = 0, then the divisor g(x) is a factor of the dividend f(x). The other factor of f(x) is the quotient Q(x).

The relation f(x) = g(x).Q(x) + R(x) or Dividend = Divisor × Quotient + Remainder is known as Division algorithm.

### Polynomial Long Division

In polynomial long division method, following steps are to be used to divide a polynomial f(x) by the other polynomial g(x):

1)  Arrange the dividend f(x) and divisor g(x) in standard form i.e. generally descending powers of variable x.
2)  Divided the first term of dividend f(x) by the first term of divisor g(x) to get the first term of quotient Q(x).
3)  Multiply each term of divisor g(x) by the first term of quotient Q(x) obtained in step 2 and subtract the product so obtained from the dividend f(x).
4)  Take the remainder obtained in step 3 as new dividend and continue the above process until the degree of the remainder is less than the divisor.

Example: Divide the polynomial x3 + 7x2 – 3 by x + 2

Solution: Here,

Quotient = x2 + 5x – 10, Remainder = 17

### Synthetic Division

Synthetic division is the process which helps us to find the quotient and the remainder when a polynomial f(x) is divided by x  a. The method of dividing a polynomial by the polynomial x – a is lengthy. Such division of polynomial can be performed in short by synthetic division method.

Study the following example of synthetic division of (3x3 – 4x2 + 7x  3) by (x – 2)

The constant term in divisor with sign changed = 2. Writing coefficients of dividend in order, we get
Quotient = 3.x2 + 2.x + 11 =3x2 + 2x + 11 and Remainder = 19

The following steps are to be used in synthetic division:

1)  List the coefficients of the terms of dividend in standard form and write the coefficient 0 for any missing term.
2)  Write the constant term of divisor with sign changed.
3)  Bring down the first coefficient of dividend.
4)  Multiply the coefficient by the constant term of the divisor with sign changed.
5)  Write the product under the next term and add.
6)  Multiply the result obtained by the same constant and write the result under the another coefficient and add.
7)  Repeat the process to get the last sum which is the remainder of the division process.

Example: Use the synthetic division to divide x4 – 7x2 + 3 by x + 2.

Solution: Here,
Dividend = x4 – 7x2 + 3 = x4 + 0.x3 – 7x2 + 0.x + 3
Divisor = x + 2 = x – (-2)
Writing the coefficient in order.

Quotient = 1.x3 + (-2).x2 + (-3).x + 6
= x3 – 2x2 – 3x + 6
Remainder = -9

### Application of Synthetic Division

Let Q(x) and R be the quotient and remainder when a polynomial f(x) is divided by binomial ax – b.

Then, f(x) = (ax – b) . Q(x) + R
= a(x – b/a) . Q(x) + R
= (x – b/a) . g(x) + R

Where, a.Q(x) = g(x)
or,       Q(x) = g(x)/a

Here g(x) and R are the quotient and remainder when f(x) is divided by (x – b/a). This result leads us to conclude that the process of synthetic division discussed earlier is also useful to find out the quotient and remainder when f(x) is divided by ax – b.

Since ax – b = a(x – b/a), we first divide f(x) by (x – b/a) to get the quotient g(x) and remainder R. Then, quotient g(x) is again divided by a to get the required quotient Q(x).

We proceed similarly to get the quotient and remainder when f(x) is divided by (ax + b).

Example: Find the quotient and remainder when 2x3 – 9x2 + 5x – 5 is divided by 2x – 3.

Solution: Let f(x) = 2x3 – 9x2 + 5x – 5
Here, 2x – 3 = 2(x – 3/2)
The constant term in (x – 3/2) with sign changed = 3/2
Writing the coefficients in order, we have
Now the required quotient = ½(2x2 – 6x – 4) = x2 – 3x – 2 and remainder = -11.

You can comment your questions or problems regarding the division of polynomials here.