Question

Find the value of m and n so that the polynomials f(x)=x^3-6x^2+mx-n is exactly divisble by (x-1) as well as (x-2)

Answer 1

Answer:

$m = 11$

$n = 6$

Step-by-step explanation:

$f(x) = {x}^{3} - 6 {x}^{2} + mx - n$

$f(1) = {1}^{3} - 6 .{1}^{2} + m.1 - n = 0 \\ \: \: then \: \: \: \\ m - n - 5 = 0 \: \: \: \: \: \: \: .......(1)$

$f(2) = {2}^{3} - 6. {2}^{2} + m.2 - n = 0 \\ then \\ 2m - n - 16 = 0 \: \: \: \: \: \: \: ........(2)$

subtracting (1) from (2) we get,

m=11

putting this value in (1) we get,

n=6

Answer 2

Values of m and n = 11 & 6

Given :

f ( x ) = x³ - 6x² + mx - n is exactly divisible by ( x - 1 ) and as well as ( x - 2 )

To Find :

Value of m & n ?

Solution :

f ( x ) = x³ - 6x² + mx - n divisible by ( x - 1 ) and same goes with ( x - 2 )

So,

f ( x ) is divisible by ( x - 1 ) and ( x - 2 )

Hence,

( x - 1 ) & ( x - 2 ) are the factors of f ( x ) .

So,

Let ,

• x - 1 = 0
• x = 1

Substitute this value of x in place of x in f ( x )

→ f ( x ) = x³ - 6x² + mx - n

→ f ( 1 ) =

→ ( 1 )³ - 6 ( 1 )² + m ( 1 ) - n = 0

Since,

→ ( x - 1 ) is a Factor , the remainder is zero .

→ 1 - 6 ( 1 ) + m - n = 0

→ 1 - 6 + m - n = 0

→ - 5 + m - n = 0

→ m = n + 5. [ Equation 1 ]

Similarly,

Let,

• x - 2 = 0
• x = 2

Substitute this value in place of x in f ( x )

→ f ( x ) = x³ - 6x² + mx - n

→ f ( 2 ) =( 2 )³ - 6 ( 2 )² + m ( 2 ) - n = 0

→ 8 - 6 ( 4 ) + 2m - n = 0

→ 8 - 24 + 2m - n = 0

→ - 16 + 2m - n = 0

Substitute the value of m from equation 1

→ - 16 + 2 ( n + 5 ) - n = 0

→ - 16 + 2n + 10 - n = 0

→ - 6 + n = 0

→ n = 6

Hence,

Hence,Value of n = 6

Same goes here,

Substitute the value of n in equation - 1

→ m = n + 5

→ m = 6 + 5

→ m = 11

Hence,

Values of m and n is 11 & 6.

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To be Noted :

In this solution Factor Theorem was used.