2.11
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, let us call dividend as f(a) {pronounced as function of a}then
f(a) = a3-6a +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.
In case of the Problem 2.10.3.2,
let us call dividend as f(x) (function of x) then
f(x) = x4-2x3+x-7
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 |
Remainder |
Remainder= to
value of function |
|
x3-6x +7 |
x+1 |
12 |
f(-1) |
|
x4-2x3+x-7 |
x+2 |
23 |
f(-2) |
|
x+1 |
x+1 |
0 |
f(-1) |
|
x-1 |
x-1 |
0 |
f(1) |
|
x+a |
x+a |
0 |
f(-a) |
|
x-a |
x-a |
0 |
f(a) |
|
x2+4x+4 |
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)
2.11 Problem 1: Find remainder of
x3+2x2-x+6 when divided by x-3
Solution:
Here f(x) = x3+2x2-x+6
and divisor is x-3
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+ 18-3+6 = 48 is the remainder
2.11 Problem 2: Find value of a if x+2 is a factor of 4x4+2x3-3x2+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 =
64-16-12-16+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) = 4x4+2x3-3x2+8x-20
f(-2) = 4*16+2(-8)-3*4 -16 -20 = 64-16-12-16-20 = 0
This proves that x+2 is a factor
of 4x4+2x3-3x2+8x-20
2.11 Summary of
learning
|
No |
Points studied |
|
1 |
If a polynomial
f(x) is divided by (x+a) then the 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) |
Additional
points:
Proof of Remainder
theorem:
Remainder Theorem:
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.
Conversely
Factor
Theorem:
(x+a) is a factor of the
polynomial f(x) if the remainder of f(-a) is equal to 0
and conversely if f(-a) = 0 then (x+a) is a factor of
f(x).
Proof:
Suppose f(-a) = 0
By remainder theorem, f(-a) is
the remainder we get on dividing f(x) by x+a. So from
the fact that f(-a)=0, it follows that the remainder
on dividing f(x) by x+a = 0.This means that x+a divides f(x) exactly: that is x+a
is a factor of f(x).
2.11 Problem
3: Check whether
7+3x is a factor of 3x3+7x
Solution:
Let f(x) = 3x3+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: 3x3+7x = x(3x2+7)
which clearly indicates that 7+3x is not a factor.
2.11 Problem
4: Factorise 6x2-11x+3
Solution:
Let f(x) = 6x2-11x+3
By factorisation method we find that the roots of the
equation f(x) = 0 are 1/3 and 3/2
Hence (x-1/3) and (x-3/2) are factors of f(x)
(x-1/3)*(x-3/2)
= x2 - (1/3)x – (3/2)x + 3/6
= x2 - (11/6)x +3/6
= (6x2-11x+3)/6.
Multiply both sides by 6 we get
f(x) = 6x2-11x+3 = 6(x-1/3)*(x-3/2)
2.11 Problem
5: If (x+1) is a
factor of x3 +2x2 - 5x – 6, find its other factors.
Hint:
Let f(x) = x3 + 2x2- 5x - 6
We also note that f(-1) =
-1+2+5-6 = 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)(x2+x-6)
By factorizing,
(x2+x-6)
= (x2+3x-2x-6)
= x(x+3)-2(x+3)
= (x+3)(x-2)
Hence
f(x) = (x+1)(x-2)(x+3)
2.11 Problem
6: If a quadratic
polynomial, when divided by (x-1), (x+1) and (x-2) leaves the remainders 2, 4
and 4 respectively, find the quadratic equation.
Solution:
Let the quadratic polynomial be f(x) = ax2+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
a-b+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 x2-x+2.
2.11 Problem
7: Given that px2+qx+6
leaves a remainder of 1 when divided by 2x+1 and 2qx2+6x+p
leaves a remainder of 2 when divided by 3x-1, find p and q.
Solution:
Let f(x) = px2+qx+6, g(x) = 2qx2+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. p-2q = -20 (On simplification) ---ŕ(1)
When g(x) is divided by 3x-1, 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