Question

Find the remainder f (x)= x^3-6x^2+2x-4 is divided by g (x)= 3x-1​

Answer 1

SOLUTION :

f(x) = x³ - 6x² + 2x - 4

and  g(x) = (3x -1)

⇒ 3x = 1

⇒ x = $\bf \frac{1}{3}$

Now,

$\bf x^{3} - 6x^{2} + 2x - 4 \\\\=( \frac{1}{3} )^{3} - 6(\frac{1}{3})^{2} + 2(\frac{1}{3} )- 4\\\\ =\frac{1}{27 }- \frac{6}{9} + \frac{2}{3} - 4\\\\=\frac{1 - 18 + 18 - 4}{27}\\\\=\frac{-3}{27}\\\\=\frac{-1}{9}$

so the remainder is $\bf \frac{-1}{9}$

Answer 2

Answer:

Remainder = (-1)/9

Step-by-step explanation:

Given,

f(x) = x³ - 6x² + 2x - 4

g(x) = 3x - 1

Now, finding the value of x.

=> 3x - 1 = 0

=> 3x = 1

=> x = 1/3

.°. x = 1/3

Now, putting the given value of x.

x³ - 6x² + 2x - 4

=> x³ - 6x² + 2x - 4

=> (1/3)³ - 6(1/3)² + 2(1/3) - 4

=> 1/27 - 6/9 + 2/3 - 4/1

=> (1 - 18 + 18 - 4) / 27

=> -3/27

=> -1/9

.°. Remainder = -1/9