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
12m^{3 }n^{5 } by 4 m^{2 }n
Solution:
Step 1: 12m^{3 }n^{5
}/ 4 m^{2 }n =
(12/4)* (m^{3 }n^{5 }/m^{2 }n)
Step 2:
12/4 = 3,
Step 3:
m^{3 }n^{5}/ m^{2 }n = m^{32 }n^{51 = }m^{ }n^{4}
_{}12m^{3 }n^{5 }/4 m^{2 }n = 3 m^{ }n^{4}
Verification:
Divisor*quotient
+ reminder = 4 m^{2 }n*3 m^{ }n^{4 }+0 =12^{ }m^{2+1
}n^{1+4} =12m^{3 }n^{5}^{ } which is dividend
2.10.1 Problem 2 : Divide 57x^{2}y^{2}z^{2} by 19xyz
Step 1 :
57x^{2}y^{2}z^{2}
/19xyz =
(57/19) * (x^{2}y^{2}z^{2})/xyz
Step 2:
57/19 =3
Step 3:
x^{2}y^{2}z^{2}/xyz
= x^{21}y^{21}z^{21 } = xyz
Thus
57x^{2}y^{2}z^{2} /19xyz = (57/19) * (x^{2}y^{2}z^{2})/xyz =3xyz
Verification:
(Divisor*Quotient) +
Remainder = (3xyz * 19xyz) +0 = (3*19)*xyz*xyz +0= 57x^{1+1}y^{1+1}z^{1+1}+0=57x^{2}y^{2}z^{2} 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 x^{2}y^{2}z^{2} /xyz which is nothing but quotient
of the variables (x^{2}y^{2}z^{2} 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
402^{3}m^{2}n^{2}603^{2}m^{2}n 804^{2}m^{3}
n^{4 }by (201^{2}m^{2})
Solution:
We
know that
402^{3}= (2x201)^{3}=
(2)^{3}x(201)^{3}, 603^{2 } = (3x201)^{2 }= (3)^{2}x(201)^{2},
804^{2 } = (4x201)^{2 }=
(4)^{2}x(201)^{2}
_{} [402^{3}m^{2}n^{2}603^{2}m^{2}n
804^{2}m^{3} n^{4}]/(201^{2}m^{2})
=[(2)^{3}*(201)^{3}
m^{2}n^{2}(3)^{2}*(201)^{2} m^{2}n
(4)^{2}*(201)^{2}m^{3} n^{4}]/(201^{2}m^{2})
= [
(2)^{3}*(201) n^{2}(3)^{2}* n (4)^{2}*m^{1}
n^{4}] =  (8*201* n^{2}9n 16mn^{4})
Verification:
Divisor*quotient
+ reminder
= (201^{2}m^{2})*[(8*201*
n^{2}+9n +16mn^{4})]+0=
= +(201^{2}m^{2})*(8*201*
n^{2} 201^{2}m^{2}*9n 201^{2}m^{2}*16mn^{4})
+0
= 8*201^{3}m^{2 }n^{2 } 9*201^{2}m^{2+2}n16*201^{2}m^{2+1}n^{4})
= 2^{3}*^{ }201^{3}m^{2}^{
}n^{2 }  3^{2} *201^{2}m^{4}n4^{2}*201^{2
}m^{3} n^{4}
= (2*201)^{3}m^{2}n^{2}(3*201)^{2}
m^{2}n (4*201)^{2} m^{3} n^{4}
= 402^{3} m^{2}n^{2}
 603^{2} m^{2}n  804^{2} m^{3} n^{4}
= dividend
2.10.2 Problem 2 :
Divide 2a^{4 }b^{3}+ 8a^{2 }b^{2 } by 2ab
Solution:
(2a^{4 }b^{3}+ 8a^{2 }b^{2})/2ab
= (2a^{4 }b^{3}/2ab) + (8a^{2 }b^{2 }/ 2ab) = a^{3
}b^{2} +4a^{ }b
Verification:
Divisor*quotient
+ reminder = 2ab*(a^{3 }b^{2} +4a^{ }b) +0= 2a^{4 }b^{3}+
8a^{2 }b^{2 } 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+x^{3}6x
(a trinomial) by x+1(a binomial)
Solution:
Note degree of dividend(x^{3}
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 x^{3} 6x+7
as x^{3} +0x^{2}6x+7 

3 
Divide
the first term of the dividend by the first term of the divisor,( x^{3}/x
= x^{2}) ^{ }Hence, x^{2 } 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 x^{2})
and write the
product(=x^{3}+ x^{2})
below the dividend 

5 
Subtract
result of step 4 from the given dividend. The result is ( x^{3} +0x^{2}
) (x^{3}+ x^{2}) =  x^{2} 

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 x^{2} 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)* (x^{2}x5)+12 = x*(x^{2}x5)
+1*(x^{2}x5)+12 = (x^{3}x^{2}5x)+ (x^{2}x5)+12
= x^{3}x^{2}+ x^{2}5xx 5+12 = x^{3}0x^{2}6x
+7 = x^{3}6x +7
which
is nothing but dividend
2.10.3 Problem 2: x^{5}
9x^{2} +12x14 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 (x^{4 },x^{3}). This is done by having their
coefficients as zero.
The dividend is re written as x^{5} +0x^{4}
+0x^{3}9x^{2} +12x14. The divisor is already in descending
order of power of x.
_{}
 
x^{5} 3x^{4}

3x^{4} +0x^{3}

3x^{4} 9x^{3}
 9x^{3} 9x^{2}
 9x^{3}
27x^{2}
 18x^{2}+12x
 18x^{2} 54x
66x14
66x198
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 =x^{5} 9x^{2}
+12x14 = 2^{5} 9*2^{2} +12*214 = 3236+2414 = 6
Divisor = x3 =23 = 1
Quotient = _{} = 2^{4} +3*2^{3}
+9*2^{2}+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 6p^{3}
19p^{2} 8p by p^{2} 4p+2
Solution:
6p+5
p^{2} 4p+2_{}
( ) 6p^{3}
24p^{2} +12p ΰ 
(1) {= 6p*(p^{2} 4p+2)}
(=) +5 p^{2}
20p ΰ (2) {subtract
(1) from given dividend}
( )
 5p^{2}
 20p+10 ΰ (3) {= 5*(p^{2}
4p+2)}
(=) 10 ΰ {subtract (3) from (2)}
Verification:
Quotient
*Divisor = (6p+5)*
(p^{2} 4p+2) = 6p* p^{2} +6p*4p+6p*2+5* p^{2}+5*4p+5*2 = 6p^{3} 24p^{2}+12p+5p^{2}20p+10
=6p^{3} 19p^{2}8p+10
Quotient *Divisor
+ Reminder = (6p^{3}
19p^{2}8p+10)10 = 6p^{3}
19p^{2}8p which is dividend
2.10.3 Problem 4:
Divide a^{5} +b^{5 }by a+b
Solution:
a+b_{}
() a^{5}+ a^{4}b
(=)
 a^{4}b+0
() a^{4}ba^{3}b^{2}
(=) a^{3}b^{2}+0
()  a^{3}b^{2}+ a^{2}b^{3}
(=)  a^{2}b^{3}+0^{}
^{ } ()
a^{2}b^{3}ab^{4}
(=)
ab^{4 }+ b^{5}
() ab^{4 }+
b^{5}
(=) 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 xa.
We shall describe this
method by taking the problem which was worked out earlier (2.10.3 Problem 2).
Divide x^{5}
9x^{2} +12x14 by x 3
Solution:
Write the dividend in its
standard form as: 1x^{5} + 0x^{4}
+ 0x^{3
} 9x^{2}
+ 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 coefficients
of the dividend (1, 0, 0, 9, 12, 14)
Write the coefficient 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 2^{nd}
row (in this case 3*1=3 in the 3^{rd}
column). Add these numbers in the 1^{st} and 2^{nd} 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
402^{3}m^{2}n^{2 } 603^{2}m^{2}n  804^{2}m^{3}
n^{4}.
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) = 402^{3}m^{2}n^{2
} 603^{2}m^{2}n  804^{2}m^{3} n^{4}
A polynomial f(x) in one
variable x is an algebraic expression of the form
f(x)
= a_{n}x^{n}+ a_{n1}x^{n1}+
a_{n2}x^{n2}+
. a_{2}x^{2}+ a_{1}x+
a_{0} = 0
where
a_{0},a_{1},a_{2},
a_{n1},a_{n} are
constants and a_{n }0
a_{0},a_{1},a_{2},
a_{n1} and a_{n }are called coefficients of
x^{0},x^{1},x^{2}
. x^{n1} and x^{n}^{ }respectively. n
is called the degree of the polynomial.
Each of a_{n}x^{n},
a_{n1}x^{n1,
.}
a_{2}x^{2}, a_{1}x^{1}, a_{0} are
called the terms of the polynomial.
Let f(x) = x^{5} 
9x^{2} + 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) =
19+1214= 10
Similarly, if we substitute
x = 1 we get f(1) =
36
f(a)
= a^{5}  9a^{2} + 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 x^{2}2x=0
Solution:
Let f(x) = x^{2}2x
We note that f(0) = 0^{2}2*0 = 0,
f(1)
= 1^{2}2 = 1
f(2)
= 2^{2}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) = x^{2}+5x+p and q(x) = x^{2}+3x+q have a common factor
then
(i)
Find the common factor
(ii) Show that (pq)^{2}= 2(3p5q)
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 xk
_{} f(k)
= k^{2}+5k+p = 0
Since xk is also a factor
of q(x)
_{} q(k)
= k^{2}+3k+q = 0
_{} k^{2}+5k+p = k^{2}+3k+q: On simplification
_{}k = (1/2)(qp)
Hence the common factor =
xk = x  (1/2)(qp)
= x + (1/2)(pq)
By substituting the value
of k in f(x) we get
((qp)/2)^{2}+5(qp)/2+p
= 0
i.e. (pq)^{2}/4+5(qp)/2+p
= 0
i.e. (pq)^{2}+10(qp)+4p
= 0
i.e. (pq)^{2}^{
}= 10p10q4p
= 6p10q
= 2(3p5q)