State and Prove Remainder Theorem and Factor Theorem

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) $\geq$ degree of p(x) and p(x) $\neq$ 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 $\geq$ 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 $\geq$ 1 by reminder theorem.

f(x) = ( x-a) . q(x) + f(a) . . . . . . . . . . . $\rightarrow$ equation ‘A ‘

1 . Suppose f(a) = 0

then equation ‘A’ $\Rightarrow$ 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 $\geq$ 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 $f \left ( \frac{-b}{a} \right )$

c) The remainder when f(x) is divided by ( ax – b) is $f \left ( \frac{b}{a} \right )$

Application of factor theorem

a) ( x – a) is a factor of xⁿ – aⁿ for any n $\in$ 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 $\neq$ 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 = $f \left ( \frac{-1}{2} \right )$

put x = -1/2 in above equation then we get

$f \left ( \frac{-1}{2} \right )$ = 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

$\Rightarrow$ – 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

$\Rightarrow$ -27 – 9p + 39 + q = 0

$\Rightarrow$ – 9p + q = -12 . . . . . . . . eq.n II

Subtract equation from II to I

$\Rightarrow$ – 8p = – 24

$\Rightarrow$ p = 3

Now the value of ‘ p substitute in equation – I then

$\Rightarrow$ q = 12 + 3 = 15

Therefore we get p = 3 and q = 15

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