Question

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

Answer 1

Answer

f(x)=x^3-6x^2+2x-4

g(x)=3x-1

● The remainder is -107/27.

Answer 2

Answer :

We have,

g(x) = 3x - 1 = 3

Therefore, by remainder theorem when f(x) is divided by

$\\ \boxed{ \sf g(x) = 3 \bigg(x - \frac{1}{3} \bigg)}$

, the remainder is equal to

$\\ \boxed{ \sf f \bigg( \frac{1}{3} \bigg)} .$

Now, f(x) = x³ - 6x² + 2x - 4

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

Hence, the required remainder :

$\\ \boxed{ \sf \frac{ - 107}{27} }$