2.10
Division of Polynomials
We have learnt the following
statement to be true for any number
Dividend = (divisor*quotient) + remainder.
The above relationship holds good for polynomials also.
2.10.1
Division of Monomial by monomial
2.10.1
Problem 1: Divide
12m3 n5 by 4 m2 n
Solution:
Step 1: 12m3 n5 / 4 m2 n
= (12/4)* (m3 n5 /m2
n)
Step 2:
12/4 = 3,
Step 3:
m3 n5/ m2 n = m3-2 n5-1
= m n4
12m3 n5 /4 m2 n = 3 m n4
Verification:
Divisor*quotient + reminder = 4 m2
n*3 m n4 +0 =12 m2+1 n1+4
=12m3 n5 which is dividend
2.10.1 Problem 2 : Divide 57x2y2z2 by 19xyz
Step 1 :
57x2y2z2 /19xyz = (57/19) * (x2y2z2)/xyz
Step 2:
57/19 =3
Step 3:
x2y2z2/xyz = x2-1y2-1z2-1
= xyz
Thus 57x2y2z2
/19xyz = (57/19) * (x2y2z2)/xyz
=3xyz
Verification:
(Divisor*Quotient) + Remainder = (3xyz * 19xyz) +0 =
(3*19)*xyz*xyz +0= 57x1+1y1+1z1+1+0=57x2y2z2 which is dividend!
We observe 3 is quotient of 57/19 which is nothing but
quotient of coefficients of monomials (57 and 19)
Similarly xyz is quotient of x2y2z2
/xyz which is nothing but quotient of the variables (x2y2z2
and xyz)
Steps to divide a monomial by monomial:
The quotient has two parts
coefficient and variable. How do we get these?
1. The coefficient of quotient of two monomials is equal
to the quotient of their coefficients
2. The variable part in the quotient of two monomials is
nothing but the quotient of the variables in the monomials
2.10.2 Division of a Polynomial by a
Monomial
2.10.2 Problem 1: Divide 4023m2n2-6032m2n
-8042m3 n4 by (-2012m2)
Solution:
We know that
4023= (2x201)3= (2)3x(201)3,
6032 = (3x201)2 =
(3)2x(201)2, 8042 = (4x201)2 = (4)2x(201)2
[4023m2n2-6032m2n
-8042m3 n4]/(-2012m2)
=[(2)3*(201)3 m2n2-(3)2*(201)2
m2n -(4)2*(201)2m3 n4]/(-2012m2)
= -[ (2)3*(201) n2-(3)2*
n -(4)2*m1 n4] = -
(8*201* n2-9n -16mn4)
Verification:
Divisor*quotient + reminder
= (-2012m2)*[-(8*201* n2+9n
+16mn4)]+0=
= +(2012m2)*(8*201* n2
-2012m2*9n -2012m2*16mn4)
+0
= 8*2013m2 n2 -9*2012m2+2n-16*2012m2+1n4)
= 23*
2013m2 n2 - 32 *2012m4n-42*2012
m3 n4
= (2*201)3m2n2-(3*201)2
m2n (4*201)2 m3 n4
= 4023 m2n2 - 6032
m2n - 8042 m3 n4
= dividend
2.10.2 Problem 2 : Divide 2a4
b3+ 8a2 b2
by 2ab
Solution:
(2a4 b3+
8a2 b2)/2ab = (2a4 b3/2ab) + (8a2
b2 / 2ab) = a3 b2 +4a b
Verification:
Divisor*quotient + reminder =
2ab*(a3 b2 +4a b) +0= 2a4 b3+
8a2 b2 which is dividend
Steps to divide a polynomial by the monomial:
1. Divide each term of the
polynomial by the monomial.
2. The partial quotients when expressed collectively
become the quotient of polynomial.
2.10.3
Division of a Polynomial by a Binomial (Long division method)
2.10.3 Problem 1: To begin with let us learn the steps of division by
dividing 7+x3-6x (a trinomial) by x+1(a binomial)
Solution:
Note
degree of dividend(x3
-6x+7) is 2
and degree of divisor(x+1) is 1
|
Step |
Procedure |
|
|
1 |
Arrange the terms of dividend and divisor in
descending order of their degrees(Already in descending order) |
|
|
2 |
If any
term of a degree is missing in dividend or divisor, add that degree with
coefficient as 0. Write
dividend x3 -6x+7
as x3 +0x2-6x+7 |
|
|
3 |
Divide
the first term of the dividend by the first term of the divisor,( x3/x
= x2) Hence, x2 is the first term of the quotient, write
this term on the top of the dividend. |
|
|
4 |
Multiply
divisor(which is x+1) by the first term of the quotient(which is x2)
and write the
product(=x3+ x2)
below the dividend |
|
|
5 |
Subtract
result of step 4 from the given dividend. The result is ( x3 +0x2
) (x3+ x2) = - x2 |
|
|
6 |
Take
term of next degree (=-6x) from the
given dividend and write it next to the result got in step 5. The
result is -x2 6x. Consider this as new dividend |
|
|
7 |
Repeat
steps 3 to 6, by taking terms from given dividend in step corresponding to step 6 |
|
|
8 |
Repeat
above procedure till the degree of reminder is less than degree of divisor |
Verification:
Divisor *Quotient +Reminder = (x+1)* (x2-x-5)+12 = x*(x2-x-5) +1*(x2-x-5)+12
= (x3-x2-5x)+ (x2-x-5)+12 = x3-x2+
x2-5x-x -5+12 = x3-0x2-6x +7 = x3-6x
+7
which is nothing but dividend
2.10.3
Problem 2: x5
-9x2 +12x-14 divided by x -3
Solution:
Though, the dividend is in descending order of power of x,
we need to have missing terms of powers of x (x4 ,x3).
This is done by having their co-efficients as zero.
The dividend is re written as x5 +0x4 +0x3-9x2
+12x-14. The divisor is already in descending order of power of x.
- | x5 -3x4
- |3x4
+0x3
- |3x4 -9x3
- |9x3 -9x2
- |9x3
-27x2
- |18x2+12x
- |18x2
-54x
-|66x-14
-|66x-198
184
Verification:
We can verify the solution by doing proper multiplication
of terms.
Since terms are big, let us verify by an alternative
method of substitution.
Let us find the results when x=2
Then
Dividend =x5 -9x2 +12x-14 = 25
-9*22 +12*2-14 = 32-36+24-14 = 6
Divisor = x-3 =2-3 = -1
Quotient = 
= 24 +3*23
+9*22+18*2+66 = 16+24+36+66=178
Quotient*Divisor + Reminder = 178*-1+184 = -178+184= 6
which is dividend
2.10.3
Problem 3: Divide 6p3 -19p2 -8p
by p2 -4p+2
Solution:
6p+5
p2
-4p+2
( -) |6p3 -24p2 +12p --ΰ ---- (1)
{= 6p*(p2 -4p+2)}
(=) |+5 p2
-20p --ΰ
-----(2) {subtract (1) from given dividend}
( -) |
5p2 - 20p+10 --ΰ
-----(3) {= 5*(p2 -4p+2)}
(=) -10
--ΰ {subtract (3) from (2)}
Verification:
Quotient *Divisor = (6p+5)* (p2 -4p+2)
= 6p* p2 +6p*-4p+6p*2+5* p2+5*-4p+5*2 = 6p3 -24p2+12p+5p2-20p+10
=6p3 -19p2-8p+10
Quotient *Divisor + Reminder = (6p3 -19p2-8p+10)-10 = 6p3 -19p2-8p which is
dividend
2.10.3
Problem 4:
Divide a5 +b5 by
a+b
Solution:
a+b
(-) |a5+
a4b
(=) - a4b+0
(-) |a4b-a3b2
(=)
a3b2+0
(-)
| a3b2+ a2b3
(=) - a2b3+0
(-) |-a2b3-ab4
(=)
ab4 + b5
(-) |ab4 + b5
(=)
0
Exercise : Verify that Divisor*Quotient+
Reminder = dividend
2.10 Summary of learning
|
No |
Points studied |
|
1 |
Division
of algebraic expressions |
Additional
points:
Synthetic method
(Horners method) of
division when the divisor is of the form x-a.
We shall describe this method by taking the problem which
was worked out earlier (2.10.3 Problem 2).
Divide x5 -9x2 +12x-14
by x -3
Solution:
Write the dividend in its standard form as: 1x5 + 0x4 + 0x3 - 9x2 + 12x - 14.
Here the constant term in the divisor is -3
First, write the negative of constant term in the divisor
(3 in this case) in the first column of the first row. In the next
columns of the first row
write the co-efficients of the dividend (1, 0, 0, -9, 12, -14)
Write the co-efficient of the first term of the divisor (in
this case 1) in the corresponding column of the third row down below (in 2nd
column).
starting from this column in the third row, write the
product of divisor (in this case 3) and the number in this column (in this case
1) in the next column of 2nd row (in this case 3*1=3 in the 3rd column). Add these
numbers in the 1st and 2nd row (in this case 0+3=3) into the corresponding
column in the third row. Repeat this process till the result in the last column
in the third row is got. The value in the last column of the third row gives
the reminder.
|
Divisor |
Dividend portion |
|
|||||
|
3 |
1 |
0 |
0 |
-9 |
12 |
-14 |
First Row |
|
|
|
3(=3*1) |
9(= 3*3) |
27(= 3*9) |
54(= 3*18) |
198(= 3*66) |
Second
row |
|
|
1 |
3=(0+3) |
9(= 0+9) |
18(=-9+27) |
66(=12+54) |
184(=-14+198) |
Third
row |
You will observe that the reminder is 184,
which is same as what we got while solving problem (2.10.3 Problem 2)
Roots of an
equation:
Let us take the polynomial 4023m2n2
- 6032m2n - 8042m3 n4.
Since this polynomial contains m and n as variables we can
denote the same by f(m,n).
f(m,n) is pronounced as function
of m and n.
f(m,n) = 4023m2n2 - 6032m2n
- 8042m3 n4
A polynomial f(x) in one variable x is an algebraic
expression of the form
f(x) = anxn+ an-1xn-1+
an-2xn-2+
. a2x2+ a1x+
a0 = 0
where a0,a1,a2,
an-1,an
are constants and an 
0
a0,a1,a2,
an-1
and an are called co-efficients
of x0,x1,x2
.
xn-1 and xn respectively. n is called the degree of the polynomial.
Each of anxn, an-1xn-1,
.
a2x2, a1x1, a0 are
called the terms of the polynomial.
Let f(x) = x5 - 9x2 + 12x - 14
If we substitute x = 0 we get f(0) = 0 -9*0 +12*0 -14 = -14
If we substitute x = 1 we get f(1) = 1-9+12-14= -10
Similarly, if we substitute x = -1 we get f(-1) = -36
f(a) = a5 - 9a2 + 12a - 14
If for any value of a (x=a), f(x) = 0, then we say that
a is a root of the equation f(x)=0.
2.10.3
Problem 5: Check
if 0, 1, 2 are roots of the equation x2-2x=0
Solution:
Let f(x) = x2-2x
We note that f(0) = 02-2*0 = 0,
f(1) = 12-2 = -1
f(2) = 22-2*2 = 0
Thus 0 and 2 are the roots of the given polynomial but 1
is not.
2.10.3
Problem 6: If
f(x) = x2+5x+p and q(x) = x2+3x+q have a common factor
then
(i) Find the common factor
(ii) Show that (p-q)2= 2(3p-5q)
Solution:
Since degree of f(x) is 2 and it has a common factor, the
degree of the factor has to be one.
Let it be x-k
f(k) = k2+5k+p
= 0
Since x-k is also a factor of q(x)
q(k) = k2+3k+q
= 0
k2+5k+p = k2+3k+q:
On simplification
k = (1/2)(q-p)
Hence the common factor = x-k = x - (1/2)(q-p)
= x + (1/2)(p-q)
By substituting the value of k in f(x) we get
((q-p)/2)2+5(q-p)/2+p = 0
i.e. (p-q)2/4+5(q-p)/2+p = 0
i.e. (p-q)2+10(q-p)+4p = 0
i.e. (p-q)2 = 10p-10q-4p
=
6p-10q
=
2(3p-5q)