Question

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

Answer 1

when we divide f(X) by g(X)

g(X) = 3x-1

so root = 1/3

then by remainder theorem

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

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

= -107/27

Answer 2

According to the question;

$f(x) = x^{3} - 6x^{2} +2x - 4\\g(x) = 3x -1$

where $f(x)$ is an equation and $g(x)$ is zero of the equation.

now

$g(x) = 3x -1$ , we can also write $3x -1 =0$

$3x -1 =0\\3x =1\\x =\frac{1}{3}$

put $x =\frac{1}{3}$ in the equation $f(x) = x^{3} - 6x^{2} +2x - 4$

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

when  $f (x)= x^3-6x^2+2x-4$ is divided by $g (x)= 3x-1$ we get the remainder $\frac{-107}{27}$.

#SPJ2