Theorem:

Letf(x)be any polynomial of degree greater than or equal to one and let ‘a‘ be any number. Iff(x)is divided by the linear polynomial(x-a)then the remainder isf(a).

Remainder Theorem Proof:

Letf(x)be any polynomial with degree greater than or equal to1.

Further suppose that whenf(x)is divided by a linear polynomialp(x) = ( x -a), the quotient isq(x)and the remainder isr(x).

In other words ,f(x)andp(x)are two polynomials such that the degree off(x)is greater then equal todegree of p(x)andp(x)is not equal 0then we can find polynomialsq(x)andr(x)such that, wherer(x) = 0or 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 whenf(x)is divided by the linear polynomial(x-a)isf(a).

Statement of Factor Theorem:

If f(x)is a polynomial of degreengreater and equal to 1and‘ a‘ is any real number then

(x -a)is a factor off(x),iff(a) = 0.

and its converse ” if(x-a)is a factor of a polynomialf(x)thenf(a) = 0“

Factor Theorem Proof:

Given thatf(x)is a polynomial of degreengreater and equal to1by reminder 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)

Which shows that( x-a)is a factor off(x).Hence proved

2 .Conversely suppose that(x-a)is a factor off(x).

This implies that f(x) = ( x-a) . q(x) for some polynomial q(x).

∴f(a) = ( a-a) . q(a) = 0.

Hencef(a) = 0when(x-a)is a factor off(x).

The factor theorem simply say that If a polynomialf(x)is divided byp(x)leaves remainder zero thenp(x)is factor off(x

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 above equation then we get

f (-1) = (-1)3– 2(-1)2+ 5(-1) + 8 = – 1 – 2 – 5 + 8 = 0

Remainder = 0