Question

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Answer 1

Question :-

Find the remainder when f ( x ) = x³ - 6x² + 2x - 4 is divided by g ( x ) = 1 - 3x

Answer :-

- 107 / 27 is the remainder

Explanation :-

Given :-

→ f ( x ) = x³ - 6x² + 2x - 4

→ g ( x ) = 1 - 3x

Finding zero of g ( x ) = 1 - 3x

⇒ 1 - 3x = 0

⇒ 1 = 3x

⇒ 1 / 3 = x

⇒ x = 1 / 3

By Remainder theorem f ( 1 / 3 ) is the remainder

Substituting x = 1 / 3 in f ( x )

f ( 1 / 3 ) = ( 1 / 3 )³ - 6( 1 / 3 )² + 2( 1 / 3 ) - 4

= ( 1³ / 3³ ) - 6( 1² / 3² ) + ( 2 / 3 ) - 4

= ( 1 / 27 ) - 6( 1 / 9 ) + ( 2 / 3 ) - 4

= ( 1 / 27 ) - 2( 1 / 3 ) + ( 2 / 3 ) - 4

= ( 1 / 27 ) - ( 2 / 3 ) + ( 2 / 3 ) - 4

= ( 1 / 27 ) - 4

= (1 - 108) / 27

= - 107 / 27

Hence, -107/ 27 is the remainder

Answer 2

SOLUTION:-

Given:

f(x)= x³ -6x² +2x -4

g(x)= 1-3x is a factor of g(x)

To find:

The remainder of the cubic polynomial.

Explanation:

g(x),1 -3x=0

g(x),1= 3x

g(x),x= 1/3.

Putting the value of g(x) in f(x), we get;

$f( \frac{1}{3} ) = ( \frac{1}{3} ) {}^{3} - 6( \frac{1}{3} ) {}^{2} + 2( \frac{1}{3} ) - 4 \\ \\ f( \frac{1}{3} ) = \frac{1}{27} - \frac{6}{9} + \frac{2}{3} - 4 \\ \\ f( \frac{1}{3} ) = \frac{1 - 18 + 18 - 108}{27} \\ \\ f( \frac{1}{3} ) = \frac{1 - 108}{27} \\ \\ f( \frac{1}{3} ) = - \frac{107}{27}$

Thus,

The remainder is -107/27.