Question

find the remainder when f(x)x^3-6x^2+2x-4 is divided by 3x-1

Answer 1

Answer :-

The remainder is -107/27 respectively.

Solution :-

f(x) = x³ - 6x² +2x - 4 divided by the linear polynomial g(x) = 3x - 1

At first, we will find out the zero of the polynomial 3x - 1

=> 3x - 1 = 0

=> 3x = 1

=> x = 1/3

From the remainder thereom, we know that f(x) when divided by g(x) gives the remainder f(1/3)

Putting the value of x

${x}^{3} - 6 {x}^{2} + 2x - 4$

$= {( \dfrac{1}{3}) }^{3} - 6 \times { (\dfrac{1}{3}) }^{2} + 2 \times \dfrac{1}{3} - 4$

$= \dfrac{1}{27} - 6 \times \dfrac{1}{9} + \dfrac{2}{3} - 4$

$= \dfrac{1}{27} - \dfrac{2}{3} + \dfrac{2}{3} - 4$

$= \dfrac{1}{27} - 4$

$= \dfrac{1 - 108}{27}$

$= - \dfrac{107}{27}$

Hence, the remainder is - 107/27.

Answer 2

Solution

$given \: p(x) \: {x}^{3 } - 6 {x}^{2} + 2x - 4$

$\: \: \: \: \: \: g(x) =3 x - 1$

By using remainder theorm

$g(x) = 0$

$\implies \: 3x - 1 = 0$

$\implies \: 3x = 0 + 1$

$\implies \: x = \frac{1}{3}$

$p ( x) = x ^{3} - 6 {x}^{2} + 2x - 4$

Putting the value of x

$\implies \: p ( \frac{1}{3} )\: = \:( { \frac{1}{3} )}^{3} - 6 \times ( { \frac{1}{3} })^{2} + 2 \times \frac{1}{3} - 4$

$\implies \: \frac{1}{27} - \cancel{6} \times \dfrac{1} {\cancel{9}} + \frac{2}{3} - 4$

$\implies \: \frac{1}{27} - \frac{2}{3} + \frac{2}{3} - 4$

$- \frac { 2}{3} \: and \: + \frac{2}{3} \: got \: cancled$

$\implies \frac{1}{27} - 4$

LCM of both the number is 27

$\therefore \: \frac{1 \times 1}{27 \times 1} - \frac{4 \times 27}{1 \times 27}$

$\implies \: \frac{1 - 108}{27}$

$= \large \underline{ \underline{ \boxed {{ - \frac{107}{27}}}}} \: answer$