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