REMAINDER THEOREM AND FACTOR THEOREM
Remainder Theorem
In this section , we shall study a simple and an elegant method of finding the remainder.
In the case of divisibility of a polynomial by a linear polynomial we use a well known theorem called Remainder Theorem.
Remainder Theorem :
If a polynomial p(x) of degree greater than or equal to one is divided by a linear polynomial (x–a) then the remainder is p(a), where a is any real number.
Significance of Remainder theorem :
It enables us to find the remainder without actually following the cumbersome process of long division.
Note :
(i) If p(x) is divided by (x+a), then the remainder is
p(– a)
(ii) If p(x) is divided by (ax–b), then the remainder is
p(b/a)
(iii) If p(x) is divided by (ax+b), then the remainder is
p(-b/a)
Example 1 :
Using Remainder Theorem, find the remainder when
f(x) = x3 + 3x2 + 3x + 1
is divided by (x + 1).
Solution :
Here, the divisor is (x + 1).
Equate the divisor to zero.
x + 1 = 0
Solve for x.
x = -1
To find the remainder, substitute -1 for x into the function f(x).
f(-1) = (-1)3 + 3(-1)2 + 3(-1) + 1
f(-1) = -1 + 3(1) - 3 + 1
f(-1) = -1 + 3 - 3 + 1
f(-1) = 0
So, the remainder is 0.
Example 2 :
Using Remainder Theorem, find the remainder when
f(x) = x3 - 3x + 1
is divided by (2 - 3x).
Solution :
Here, the divisor is (2 - 3x).
Equate the divisor to zero.
2 - 3x = 0
Solve for x.
-3x = -2
x = 2/3
To find the remainder, substitute 2/3 for x into the function f(x).
f(2/3) = (2/3)3 - 3(2/3) + 1
f(2/3) = 8/27 - 2 + 1
f(2/3) = 8/27 - 1
f(2/3) = 8/27 - 27/27
f(2/3) = (8 - 27)/27
f(2/3) = -19/27
So, the remainder is -19/27.
Example 3 :
For what value of k is the polynomial
2x4 + 3x3 + 2kx2 + 3x + 6
is divisible by (x + 2).
Solution :
Let
f(x) = 2x4 + 3x3 + 2kx2 + 3x + 6
Here, the divisor is (x + 2).
Equate the divisor to zero.
x + 2 = 0
Solve for x.
x = -2
To find the remainder, substitute -2 for x into the function f(x).
f(-2) = 2(-2)4 + 3(-2)3 + 2k(-2)2 + 3(-2) + 6
f(-2) = 2(16) + 3(-8) + 2k(4) - 6 + 6
f(-2) = 32 - 24 + 8k - 6 + 6
f(-2) = 8 + 8k
So, the remainder is (8 + 8k).
If f(x) is exactly divisible by (x + 2), then the remainder must be zero.
Then,
8 + 8k = 0
Solve for k.
8k = -8
k = -1
Therefore, f(x) is exactly divisible by (x+2) when k = –1.
Factor Theorem
If p(x) is a polynomial of degree n ≥ 1 and ‘a’ is any real number then
(i) p(a) = 0 implies (x - a) is a factor of p(x).
(ii) (x - a) is a factor of p(x) implies p(a) = 0.
Note :
(i) (x - a) is a factor of p(x), if p(a) = 0.
(ii) (x + a) is a factor of p(x), if p(-a) = 0.
(iii) (ax + b) is a factor of p(x), if p(-b/a) = 0.
(iv) (x - a)(x - b) is a factor of p(x), if
p(a) = 0 and p(b) = 0
Example 1 :
Using Factor Theorem, show that (x + 2) is a factor of
x3 - 4x2 - 2x + 20
Solution :
Let
f(x) = x3 - 4x2 - 2x + 20
Equate the factor (x + 2) to zero.
x + 2 = 0
Solve for x.
x = -2
By Factor Theorem,
(x + 2) is factor of f(x), if f(-2) = 0
Then,
f(-2) = (-2)3 - 4(-2)2 - 2(-2) + 20
f(-2) = -8 - 4(4) + 4 + 20
f(-2) = -8 - 16 + 4 + 20
f(-2) = 0
Therefore, (x + 2) is a factor of x3 - 4x2 - 2x + 20.
Example 2 :
Is (3x - 2) a factor of 3x3 + x2 - 20x + 12 ?
Solution :
Let
f(x) = 3x3 + x2 - 20x + 12
Equate the factor (3x + 2) to zero.
3x - 2 = 0
Solve for x.
3x = 2
x = 2/3
By Factor Theorem,
(3x - 2) is factor of f(x), if f(2/3) = 0
Then,
f(2/3) = 3(2/3)3 + (2/3)2 - 20(2/3) + 12
f(2/3) = 3(8/27) + 4/9 - 40/3 + 12
f(2/3) = 8/9 + 4/9 - 40/3 + 12
f(2/3) = 8/9 + 4/9 - 120/9 + 108/9
f(2/3) = (8 + 4 - 120 + 108) / 9
f(2/3) = (120 - 120) / 9
f(2/3) = 0
Therefore, (3x - 2) is a factor of 3x3 + x2 - 20x + 12.
Example 3 :
Find the value of m, if (x - 2) is a factor of the polynomial
2x3 - 6x2 + mx + 4
Solution :
Let
f(x) = 2x3 - 6x2 + mx + 4
Equate the factor (x - 2) to zero.
x - 2 = 0
Solve for x.
x = 2
By Factor Theorem,
(x - 2) is factor of f(x), if f(2) = 0
Then,
f(2) = 0
2(2)3 - 6(2)2 + m(2) + 4 = 0
f(2) = 2(8) - 6(4) + 2m + 4 = 0
f(2) = 16 - 24 + 2m + 4 = 0
f(2) = 2m - 4 = 0
2m = 4
m = 2
Therefore (x - 2) is a factor of f(x), when m = 2.
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