2.11. Remainder Theorem:
2.11.1 Roots
of an equation/Zero of a polynomial
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.
If in case of polynomial f(x), if f(a)=0, then
'a' is called 'zero' of polynomial f(x).
2.11.2 More
on divisions:
We know that like in case of numbers, following relationship holds good in case of polynomials
also.
Divisor*quotient
+ reminder =^{ } dividend
Above relationship is called
f(x)= g(x)*q(x)+ r(x) (1)
[
Here the divisor g(x)divides f(x) to give the quotient q(x) and reminder
r(x). Note g(x)_{} 0 and r(x) =0 or its degree
is less than degree of g(x) ]
Note
in (1) if any three polynomials are given then
the fourth polynomial can be found
If f(x), q(x) and r(x)
are given then g(x)= {f(x)r(x)}/q(x)
If f(x), g(x) and q(x)
are given then r(x)= f(x){ g(x) *q(x)}
Can we find quotient and reminder on
dividing a polynomial without the actual division?
2.11 Problem 1: Find the quotient and reminder on dividing f(x) =
x^{3}+4x^{2}6x+2 by g(x)= (x3)
Solution:
We know f(x)= g(x)*q(x)+ r(x) and
also degree of r(x) is less than that of g(x). Thus degree of g(x) has to be 0.
Since degree of dividend is 3 and degree of divisor is 1, degree of quotient
has to be 2(=31).
Hence let r(x) be = k a
constant and q(x) = ax^{2}+bx+c for some value of a, b and c.
_{} x^{3}+4x^{2}6x+2
=(x3)* (ax^{2}+bx+c) = (ax^{3}+bx^{2}+cx)+(3ax^{2}3bx3c)+
k = ax^{3}+x^{2}(b3a)+x(c3b)+k3c.
_{}a=1;4=b3a; 6=c3b;2=k3c ( Equating coefficents of
respective terms)
Thus a=1; b=4+3a;
c=3b6; k=2+3c
By simplifying further
we get a=1, b=7, c=15 and k= 47
_{} q(x) = ax^{2}+bx+c=
q(x) = x^{2}+7x+15 and r(x) = 47
Verification:
Verify that x^{3}+4x^{2}6x+2
= (x3)* (x^{2}+7x+15)+47
How to modify a polynomial so as to be exactly divisible by
another polynomial?
We know that
f(x)= g(x)*q(x)+ r(x)
if g(x) has to divide
f(x) exactly, then r(x) has to be zero
·
First divide f(x) by g(x) by any of the method( long
division as learnt in the lesson 2.10 or by the method mentioned in problem
2.11.1) to get the reminder r(x).
·
Subtract this r(x) from f(x) to get a new polynomial which is divisible exactly by g(x).
2.11 Problem 2:
What should be added to or subtracted
from x^{3}+5x^{2}+5x+8 to make it exactly divisible by x^{2}+3x2?
Solution:
When we divide x^{3}+5x^{2}+5x+8
by x^{2}+3x2 we get the reminder as 1060
Since for exact
divisibility reminder has to be zero, we need to subtract this reminder from
the dividend and hence the answer is (x^{3}+5x^{2}+5x+8)
– (1060)= x^{3}+5x^{2}+5x+1060
2.11.2 Remainder
Theorem:
We notice that the method mentioned in section 2.10
to find the remainder by repeated process of divisions, is a long process and
is time consuming..
Is there a simple method by which remainder can be
found?
By analyzing the problems solved in Section 2.10 we
can observe a few things.
In case of the Problem 2.10.3.1 where we divided (x^{3} 6x+7) by (x+1)
let us call dividend as f(a) {pronounced as
function of a}then
f(a) = a^{3}6a +7
Let us find the values of f(a) for various values
of a (say 1, 2,0,1,2)
Then we have
f(1) = 2, f(0) =7, f(1) = 12, f(2) = 11.
What do we observe? We notice f(1)=12 is the remainder.
Let us take the problem of dividing of
(x^{4}2x^{3}+x7)
by (x+2)
let us call dividend as f(x)
(function of x) then
f(x) = x^{4}2x^{3}+x7
Let us find the value of f(x) for
various vales of x (say 1, 2, 0,1,2)
Then we have
f(1) = 6, f(2)
=5, f(0) =7, f(1) =7, f(2)=23 is the remainder
Let us do some
tabulation of remainders for various values of dividends and divisors
Dividend
f(x) 
Divisor g(x) 
Quotient q(x) 
Remainder r(x) 
Remainder=
value of function f(k) 
x^{3}6x +7 
x+1 
x^{2}x5 
12 
f(1) 
x^{4}2x^{3}+x7 
x+2 
x^{3}4x^{2}+8x15 
23 
f(2) 
x+1 
x+1 
1 
0 
f(1) 
x1 
x1 
1 
0 
f(1) 
x+a 
x+a 
1 
0 
f(a) 
xa 
xa 
1 
0 
f(a) 
x^{2}+4x+4 
x+2 
x+2 
0 
f(2) 
We conclude that if a polynomial
f(x) is divided by a monomial of type (x+a) then the remainder is equal to
f(a)
This is called Remainder
Theorem:
If a Polynomial f(x),
over the set of real numbers R is
divided by (x+a) then the remainder is
f(a).
Note: “over the set of real
numbers” means that x can take the value of any real number (x _{} real number)
Proof of
Remainder theorem:
Remainder theorem states that if a
Polynomial f(x), is divided by (x+a) then the remainder is f(a).
Let q(x) and r(x) be the quotient and the remainder
when we divide f(x) by x+a.
Proof:
We have Dividend = Quotient*Divisor + Remainder
_{} f(x) = q(x)*(x+a) +
r(x)
Note that the degree of the divisor (=(x+a)) is 1.
Also note that the degree of remainder (= r(x)) has
to be less than the degree of divisor.
Hence, the degree of remainder = 0 which means that
r(x) does not contain the term x and so it has to be a constant (say ‘r’).
_{}f(x) = q(x)*(x+a)+r
Since the above equation is true for any value of
x, it has to be true for x = a also.
_{}f(a) = q(a)*(a+a)+r = q(a)*0+r = r
This proves the theorem.
Note: If a Polynomial f(x), is divided by (ax+b)
then the remainder is f(b/a) [ _{} ax+b = (x+b/a) ]
Factor Theorem:
If f(a) is
equal to 0 then (x+a) is a factor of the
polynomial f(x)
Proof:
Suppose f(a) = 0
By remainder theorem, on dividing f(x) by (x+a), the
remainder is f(a). So from the fact that f(a)=0, it follows that the
remainder on dividing f(x) by x+a is 0. This means that x+a divides f(x)
exactly. That is x+a is a factor of f(x).
2.11 Problem 3: Find remainder of x^{3}+2x^{2}x+6 when
divided by x3
Solution:
Here f(x) = x^{3}+2x^{2}x+6
and divisor is x3
As per the remainder theorem, if
divisor is of the form x+a then the remainder is f(a).
Since divisor is of the form x+a ,
f((3) = f(3) is the remainder
Substituting 3 in f(x), we get
f(3) = 27+ 183+6 = 48 is the remainder
2.11 Problem 4: Find value of a if x+2 is a factor
of 4x^{4}+2x^{3}3x^{2}+8x+5a
Solution:
Since x+2 is a factor of f(x),
remainder has to be 0.
Thus by Remainder theorem f(2) =0 ^{}
f(2) = 4*16+2*(8)3*4 16+5a =
64161216+5a = 20 +5a
Since f(2) =0 we have 20+5a = 0
i.e. 5a = 20 i.e. a= 4
Verification:
Since a= 4, substitute 4 for a
in f(x)
We get f(x) = 4x^{4}+2x^{3}3x^{2}+8x20
f(2) = 4*16+2(8)3*4 16 20 = 6416121620 = 0
This proves that x+2 is a factor
of 4x^{4}+2x^{3}3x^{2}+8x20
2.11 Problem 5: Check whether 7+3x is a factor of 3x^{3}+7x
Solution:
Let f(x) = 3x^{3}+7x
If 7+3x has to be a factor of f(x), then 3*(7/3+x) has to be necessarily a factor of f(x) (If y is a factor of f(x) then ny/n has
to be a factor of f(x) as y*f(x) = (ny/n)*f(x) when n is any non zero number)
_{}f(7/3) = 3(7/3)^{3 }+7(7/3) = 343/9 49/3 _{}0
Thus 7+3x is not a factor of the given polynomial.
Note: 3x^{3}+7x = x(3x^{2}+7) which
clearly indicates that 7+3x is not a factor.
2.11 Problem 6: 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.
2.11 Problem 7: 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)
2.11 Problem 8: Factorise 6x^{2}11x+3 given that f(1/3) and f(3/2)=0
Solution:
Let f(x) = 6x^{2}11x+3
It is given
that (x1/3) and (x3/2) are factors of
f(x)
(x1/3)*(x3/2)
= x^{2 } (1/3)x – (3/2)x + 3/6
= x^{2 } (11/6)x +3/6
= (6x^{2}11x+3)/6.
Multiply both sides by 6 we get
_{}f(x) = 6x^{2}11x+3 = 6(x1/3)*(x3/2)
2.11 Problem 9: If (x+1) is a factor of x^{3 }+2x^{2 }
5x – 6, find its other factors.
Hint:
Let f(x) = x^{3 }+ 2x^{2} 5x  6
We also note that f(1) = 1+2+56 = 0 thus clearly
(x+1) is a factor of f(x).
Using the division method learnt in section 2.10,
we find that
f(x) = (x+1)(x^{2}+x6)
By factorizing,
(x^{2}+x6)
= (x^{2}+3x2x6)
= x(x+3)2(x+3)
= (x+3)(x2)
Hence
f(x) = (x+1)(x2)(x+3)
2.11 Problem 10: If a quadratic polynomial, when divided by (x1),
(x+1) and (x2) leaves the remainders 2, 4 and 4 respectively, find the
quadratic equation.
Solution:
Let the quadratic polynomial be f(x) = ax^{2}+bx+c
From the given data f(1)=2, f(1)=4 and f(2)=4
But f(1) = a+b+c, f(1) = a–b+c and f(2) = 4a+2b+c
_{} We have
a+b+c = 2
ab+c = 4
4a+2b+c = 4
When we solve these three simultaneous equations as
explained in 2.14.3 we get
a=1, b=1 and c=2
Hence the quadratic equation is x^{2}x+2.
2.11 Problem 11: Given that px^{2}+qx+6 leaves a remainder
of 1 when divided by 2x+1 and 2qx^{2}+6x+p leaves a remainder of 2 when
divided by 3x1, find p and q.
Solution:
Let f(x) =
px^{2}+qx+6, g(x) = 2qx^{2}+6x+p
When f(x) is divided by 2x+1, the remainder is 1.
This implies that f(1/2) = 1
_{} p/4 –q/2+6 = 1
i.e. p2q = 20 (On simplification) ŕ(1)
When g(x) is divided by 3x1, the remainder is 2.
This implies that g(1/3) = 2
_{} 2q/9 + 6/3 +p = 2
i.e. 9p+2q = 0 (On simplification) ŕ(2)
On solving (1) and (2) we get
p = 2 and q = 9
Synthetic
method (Horner’s method) of division when the divisor is of the form xa.
Using this method we can
find quotient and reminder easily compared to long division method.
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
1.
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)
2.
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).
3.
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).
4.
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.
5.
Repeat this process till the result in the last column in
the third row is got.
6.
The value in the last column of the third row gives the
reminder.
7.
Other values in the third row represent the coefficient of the quotient.
Divisor 
Dividend portion(Columns from 2 onwards) 


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 quotient is
1x^{4}+3x^{3}+9x^{2}+18x and reminder is 184, which is same as what we got while solving problem
(2.10.3 Problem 2)
2.11 Summary of
learning
No 
Points studied 
1 
If a
polynomial f(x) is divided by (x+a) then remainder
= f(a) for any real value of x 
2 
If a
polynomial f(x) is divided by the divisor of the form
x+a and the remainder is zero, then x+a
is a factor
of f(x) 
3 
If
f(a) is equal to 0 then (x+a) is a
factor of the polynomial f(x) 