# State and Prove Remainder Theorem and Factor Theorem | Polynomials

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## Remainder and Factor Theorem Proof |  Remainder and Factor Theorem Test

In this page given definition and proof for Remainder Theorem and Factor Theorem and also provided application of 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 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 1 by 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 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 simply say that If a polynomial f(x) is divided by p(x) leaves remainder zero then p(x) is 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 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 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 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 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 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 equation from II to I – 8p = – 24 p = 3

Now the value of ‘ p substitute in equation –  I  then q  = 12 + 3 = 15

Therefore we get p = 3 and q = 15

Related Topics :

Remainder Theorem Aptitude Exercise – I

Remainder Theorem Aptitude Exercise – II

Finding Last Digit of any Number with Power

Shortcut to Find Number of Factors of a Number

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