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 x*^{3}+ 7x^{2}– 3 by x + 2

*Solution:*

*Here,*

*∴*

*Quotient = x*^{2}+ 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 (3x

^{3}– 4x^{2}+ 7x – 3) by (x – 2)
The constant term in divisor with sign
changed = 2. Writing coefficients of dividend in order, we get

∴ Quotient = 3.x

^{2}+ 2.x + 11 =3x^{2}+ 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 x*^{4}– 7x^{2}+ 3 by x + 2.

*Solution:*

*Here,*

*Dividend = x*^{4}– 7x^{2}+ 3 = x^{4}+ 0.x^{3}– 7x^{2}+ 0.x + 3

*Divisor = x + 2 = x – (-2)*

*Writing the coefficient in order.*

*Quotient = 1.x*^{3}+ (-2).x^{2}+ (-3).x + 6

*= x*^{3}– 2x^{2}– 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 2x*^{3}– 9x^{2}+ 5x – 5 is divided by 2x – 3.

*Solution:*

*Let f(x) = 2x*^{3}– 9x^{2}+ 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(2x*^{2}– 6x – 4) = x^{2}– 3x – 2 and remainder = -11.*You can comment your questions or problems regarding the*

**division of polynomials**here.
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