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/Trinomial (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)