Remainder and Factor Theorem Proof | Remainder and Factor Theorem Test
This page gives a definition and proof for the Remainder Theorem and Factor Theorem and also provides an application of the remainder theorem and factor theorem
Statement of Remainder Theorem:
Let f(x) be any polynomial of degree greater than or equal to one and let ‘ a‘ be any number. If f(x) is divided by the linear polynomial (x-a) then the remainder is f(a).
Remainder Theorem Proof:
Let f(x) be any polynomial with a degree greater than or equal to 1.
Further suppose that when f(x) is divided by a linear polynomial p(x) = ( x -a), the quotient is q(x) and the remainder is r(x).
In other words, f(x) and p(x) are two polynomials such that the degree of f(x) ≥ degree of p(x) and p(x) ≠ 0 then we can find polynomials q(x) and r(x) such that, where r(x) = 0 or degree of r(x) < degree of g(x).
By division algorithm
f(x) = p(x) . q(x) + r(x)
∴ f(x) = (x-a) . q(x) + r(x) [ here p(x) = x – a ]
Since degree of p(x) = (x-a) is 1 and degree of r(x) < degree of (x-a)
∴ Degree of r(x) = 0
This implies that r(x) is a constant, say ‘ k ‘
So, for every real value of x, r(x) = k.
Therefore, f(x) = ( x-a) . q(x) + k
If x = a,
then f(a) = (a-a) . q(a) + k = 0 + k = k
Hence, the remainder when f(x) is divided by the linear polynomial (x-a) is f(a).
Statement of Factor Theorem:
If f(x) is a polynomial of degree n ≥1 and ‘ a ‘ is any real number, then
1. (x -a) is a factor of f(x), if f(a) = 0.
2. and its converse ” if (x-a) is a factor of a polynomial f(x) then f(a) = 0 “
Factor Theorem Proof:
Given that f(x) is a polynomial of degree n ≥ 1 by the remainder theorem.
f(x) = ( x-a) . q(x) + f(a) . . . . . . . . . . . ⇒ equation ‘A ‘
1 . Suppose f(a) = 0
then equation ‘A’ ⇒ f(x) = ( x-a) . q(x) + 0 = ( x-a) . q(x)
This shows that ( x-a) is a factor of f(x). Hence proved
2 . Conversely, suppose that (x-a) is a factor of f(x).
This implies that f(x) = ( x-a) . q(x) for some polynomial q(x).
∴ f(a) = ( a-a) . q(a) = 0.
Hence, f(a) = 0 when (x-a) is a factor of f(x).
The factor theorem says that if a polynomial f(x) is divided by p(x) and leaves the remainder zero, then p(x) is a factor of f(x)
Application of remainder theorem
Suppose f(x)is a polynomial of degree ≥ 1 then
a) The remainder when f(x) is divided by ( x+a) is f(-a)
b) The remainder when f(x) is divided by ( ax +b) is 
c) The remainder when f(x) is divided by ( ax – b) is 
Application of factor theorem
a) ( x – a) is a factor of xn – an for any n N (any natural number)
b) ( x + a) is a factor of xn – an if n is an even number
c) ( x + a) is a factor of xn + an if n is odd number
Application of factor theorem for quadratic polynomials:
Consider a quadratic polynomial ax2 + bx + c ( where a ≠ 0 and a,b & c are constants )
Now let its factors be (px + q) & (rx + s)
Then ax2 + bx + c = (px + q) (rx + s) = prx2 + (ps + qr)x + qs.
Comparing the coefficients of x2, x, and constants, we get that
-
- a = pr
- b = ps + qr
- c = qs
This shows that ‘ b ‘ is the sum of two numbers, ‘ ps ‘ & ‘ qr ‘ .
Whose product is (ps) (qr) = pr x qs = ac
Therefore, to factorise ax2 + bx + c , we have to write ‘ b ‘ as the sum of two numbers whose product is ‘ ac ‘
Remainder and Factor Theorem Practice
Example – 1 : Find the remainder when x3 – 2x2 + 5x + 8 is divided by x + 1
Solution: Let f(x) = x3 – 2x2 + 5x + 8
Now, divisor = x + 1 , its zero is ‘ -1 ‘
By the remainder theorem, the required remainder = f( -1)
put x = -1 in the above equation, then we get
f (-1) = (-1)3 – 2(-1)2 + 5(-1) + 8 = – 1 – 2 – 5 + 8 = 0
Remainder = 0
Example – 2 : Find the remainder when 2x3 + 4x2 – 8x + 4 is divided by 2x + 1
Solution: Let f(x) =2x3 + 4x2 – 8x + 4
Now divisor = 2x + 1 , its zero is ‘ -1/2 ‘
By the remainder theorem, the required remainder = 
put x = -1/2 in the above equation, then we get
= 2(-1/2)3 + 4(-1/2)2 – 8(-1/2) + 4 = -(1/4) + 1 + 4 + 4 = 39/4
Remainder = 39/4
Example – 3 : Find the remainder when x3 – px2 + 6x – p is divided by (x -p)
Solution: Let f(x) = x3 – px2 + 6x – p
Now divisor q(x) = x -p, its zero is ‘ p ‘
According to remainder theorem, when f(x) is divided by q(x) is f(p)
put x = p in the above equation, then we get
f(p) = p3 – p .p2 + 6p – p = 5p
Remainder = 5p.
Example – 4 : Determine whether (x +2) is a factor of x3 – 4x2 – 4x + 16
Solution: Let f(x) = x3 – 4x2 – 4x + 16
According to factor theorem, ( x+ 2) will be factor of f(x) if f(-2) = 0
put x = -2 in the above equation then we get
f(-2) = (-2)3 – 4(-2)2 – 4(-2) + 16 = – 8 – 16 + 8 + 16 = 0
Hence, ( x+2 ) is a factor of x3 – 4x2 – 4x + 16 .
Example – 5 : Find the values of p & q such that the polynomial x3 – px2 – 13x + q has ( x – 1) and ( x + 3 ) as factors.
Solution: Let f(x) = x3 – px2 – 13x + q
Since ( x – 1) is a factor of f(x). This implies f(1) = 0
i.e (1)3 – p(1)2 – 13(1) + q = 0
⇒ – p + q = 12 . . . . . . . . eq.n I
Also, since (x +3) is a factor of f(x).
This implies f(-3) = (-3)3 – p(-3)2 – 13(-3) + q = 0
⇒ -27 – 9p + 39 + q = 0
⇒ – 9p + q = -12 . . . . . . . . eq.n II
Subtract the equation from II to I
⇒ – 8p = – 24
⇒ p = 3
Now the value of ‘ p substitute in the equation – I then
⇒ q = 12 + 3 = 15
Therefore we get p = 3 and q = 15
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