Question

find the value of m and in so that the polynomial f(x)=x^3-6x^2+mx-n is axactly dividible by (x-1) as well as (x-2)​

Answer 1

Question : -

Find the value of m and n so that the polynomial f ( x ) = x³ - 6x² + mx - n is exactly divisible by ( x - 1 ) and as well as ( x - 2 )

Answer : -

Given : -

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

Required to find : -

• Value of m & n ?

Theorem used : -

Factor theorem

Solution : -

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

Since,

f ( x ) is exactly divisible by ( x - 1 ) and ( x - 2 ) we can say that ;

( 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 $\longrightarrow{\rm{\orange{Equation - 1 }}}$

Consider this as 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,

• Value of n = 6

Similarly,

Substitute the value of n in equation - 1

=> m = n + 5

=> m = 6 + 5

=> m = 11

Hence,

• Value of m = 11

Therefore,

Values of ' m ' and ' n ' = 11 & 6

Verification : -

Substitute the values of m and n in f ( x )

f ( x ) = x³ - 6x² + mx - n

f ( x ) = x³ - 6x² + 11x - 6

Now,

Let's divide of f ( x ) with ( x - 1 )

x - 1 = 0

x = 1

f ( 1 ) =

( 1 )³ - 6 ( 1 )² + 11 ( 1 ) - 6 = 0

1 - 6 + 11 - 6 = 0

12 - 12 = 0

0 = 0

LHS = RHS

Hence Proved !

Answer 2

$\mathcal{\huge{\fbox{\red{Question\:?}}}}$

Find the value of m and n so that the polynomial f ( x ) = x³ - 6x² + mx - n is exactly divisible by ( x - 1 ) and as well as ( x - 2 )

$\mathcal{\huge{\fbox{\green{AnSwEr:-}}}}$

✏ m = 11 & n = 6

$\mathcal{\huge{\fbox{\purple{Solution:-}}}}$

Given : -

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

To find : -

• Value of m & n .

Calculation :-

According to the question,

• Using Factor theorem

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

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

Now,

↗x - 1 = 0

↗x = 1

&

↗x - 2 = 0

↗x =2

✴ Putting value of x from above in x of f ( x ).

f ( x ) = x³ - 6x² + mx - n

➡ f ( 1 ) = ( 1 )³ - 6 ( 1 )² + m ( 1 ) - n = 0

➡ 1 - 6 ( 1 ) + m - n = 0

➡ 1 - 6 + m - n = 0

➡ - 5 + m - n = 0

➡ m = n + 5 ..........(1)

Again,

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.........(2)

From 1,

➡ - 16 + 2 ( n + 5 ) - n = 0

➡ - 16 + 2n + 10 - n = 0

➡ - 6 + n = 0

➡ n = 6........(3)

Putting 3 in 1,

➡> m = n + 5

➡> m = 6 + 5

➡> m = 11..........(4)

So, The value of m = 11 & n = 6.

_____________________________________

$\mathcal{\huge{\fbox{\pink{Verification:-}}}}$

f ( x ) = x³ - 6x² + mx - n

From 3 & 4 ,

➡ f ( x ) = x³ - 6x² + 11x - 6

Using factor, (x-1)

↗x - 1 = 0

↗x = 1

➡ f ( 1 ) = ( 1 )³ - 6 ( 1 )² + 11 ( 1 ) - 6 = 0

= 1 - 6 + 11 - 6 = 0

= 12 - 12 = 0

= 0 = 0

LHS = RHS

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