Dividend = ( Divisor × Quotient ) + Remainder. 1. Division Algorithm for Polynomials 2. Synthetic Division

**Division of Polynomials**

Let us consider
the numbers 13 and 5. When 13 is divided by 5 what is the quotient and remainder.?

Yes,
of course, the quotient is 2 and the remainder is 3. We write 13 = (5×2)+3

Let
us try.

Dividend = ( Divisor × Quotient ) + Remainder.

From
the above examples, we observe that the remainder is less than the divisor.

** **

Let
*p*(*x*) and *g*(*x*) be two polynomials such that degree of
*p*(*x*) __>__ degree of *g*(*x*) and *g*(*x*)*
*≠ 0.

Then
there exists unique polynomials *q*(*x*) and *r*(*x*) such that

* p*(*x*)
= *g* (*x* ) × *q* (*x* ) + *r* (*x* ) … (1)

where
*r*(*x*) = 0 or degree of *r*(*x*) < degree of *g*(*x*).

The
polynomial *p*(*x*) is the Dividend, *g*(*x*) is the Divisor,
*q*(*x*) is the Quotient and *r*(*x*) is the Remainder. Now
(1) can be written as

Dividend
= ( Divisor × Quotient ) + Remainder.

If *r*(*x*) is zero, then we say *p*(*x*) is
a multiple of *g*(*x*). In other words, *g*(*x*) divides *p*(*x*).

If it
looks complicated, don’t worry! it is important to know how to divide polynomials,
and that comes easily with practice. The examples below will help you.

**Example 3.32**

Divide** ***x*^{3}** − **4*x*^{2}** **+** **6*x*** **by** ***x*, where ,** ***x*** **≠ 0

*Solution*

Find the quotient and the remainder when** **(5*x*^{2}** − **7*x*** **+** **2)** **÷** **(*x*** − **1)

(5*x*^{2}
− 7*x* + 2) ÷ (*x* − 1)

∴
Quotient = 5*x*–2

Remainder
= 0

Find quotient and the remainder when *f*(*x*) is divided
by *g*(*x*)

(i) *f*(*x*) = (8*x ^{3}*–6

(ii) *f*(*x*) = *x*^{4} –3*x*^{3}
+ 5*x*^{2} –7, *g*(*x*) = *x*^{2} + *x*
+ 1

(i) *f*(*x*) = (8*x ^{3}*–6

(ii) *f*(*x*) =* x*^{4}* *–3*x*^{3}*
*+ 5*x*^{2}* *–7,* g*(*x*) =* x*^{2}*
*+* x *+ 1* *

** **

Synthetic Division is a shortcut method of polynomial division. The advantage of synthetic division is that
it allows one to calculate without writing variables, than long division.

Find
the quotient and remainder when *p*(*x* ) =
(3*x* ^{3} −2*x* ^{2} –
5 + 7*x*)
is divided by *d*(*x*) = *x* + 3 using synthetic division.

**Step 1**

Arrange dividend
and the divisor in standard form.

3*x*
^{3} − 2*x*^{2} + 7*x* −5 (standard form of dividend)

*x
*+*
*3 (standard form of divisor)

Write
the coefficients of dividend in the first row. Put ‘0’ for missing term(s).

3 −2 7 −5 (first row)

**Step 2** Find out the zero of the divisor.

* x *+
3 = 0 implies* x *= −3

**Step 3** Write the zero of divisor in front of
dividend in the first row. Put ‘0’ in the first column of second row.

**Step 4 **Complete the second row and third row
as shown below.

All
the entries except the last one in the third row are the coefficients of the quotient.

Then
quotient is 3*x* ^{2} −11*x* + 40 and and
remainder is -125.

Find
the quotient and remainder when (3*x* ^{3 − }4*x*^{2}
− 5) is divided^{ }by (3*x*+1) using synthetic division.^{}

Let
*p*(*x*) = 3*x*^{3} − 4*x*^{2} − 5, *d*(*x*)
=
(3*x* +1)

Standard
form: *p*(*x*) = 3*x* ^{3} −
4*x*^{2} + 0*x* −5

and
*d* (*x*) = 3*x* +1

If the
quotient on dividing *x*^{4} + 10*x*^{3} +
35*x*^{2} + 50*x* +
29 by (*x* + 4) is *x*^{3} −
*ax*^{2} +*bx* + 6 , then find the value of *a*,
*b* and also remainder.

Let*
p*(*x*)* *=* x*^{4}* *+10*x*^{3}* *+*
*35*x*^{2}* *+*
*50*x *+*
*29

Standard
form =*
x*^{4}* *+10*x*^{3}* *+*
*35*x*^{2}* *+*
*50*x *+*
*29

quotient
*x* ^{3} + 6*x*^{2} +11*x*
+
6 is compared with given quotient *x* ^{3} −
*ax* ^{2} +*bx* + 6

coefficient
of *x*^{2} is 6 = −*a* and coefficient of *x* is 11 =
*b*

Therefore,
*a* = −6 , *b* =
11 and remainder = 5 .

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