Question

please find these by using remainder method and verify by actual division​

Answer 1

$\rm \: We \: have \: \green{ ƒ(x) = 3 {x}^{4} +2 {x}^{3} - \frac{ {x}^{2} }{3} - \frac{x}{9} + \frac{2}{27} , g(x) = x+ \frac{2}{3}}$

Therefore, by remainder theorem when f(x) is divided by g(x) = x- (-2/3), the remainder is equal to f(-2/3)

$\boxed{ \underline{ \underline{ \rm \huge \green{Solution : - }}}}$

$\sf \: Now \: \green{ ƒ(x) = 3 {x}^{4} + {2x}^{3} + \frac{ {x}^{2} }{3} - \frac{x}{9} + \frac{2}{27} }$

$\tt \implies \green{ ƒ \bigg( - \frac{2}{3} \bigg) }= 3x \bigg( \frac{ - 2}{3} { \bigg)}^{4} + 2 { \bigg( \frac{ - 2}{3} \bigg)}^{3} - \frac{( \frac{ - 2}{3} {)}^{2} }{3} - \frac{( \frac{ - 2}{3}) }{9} + \frac{2}{27}$

$\implies \: \cancel3 \times \frac{16}{ \cancel{81}} + 2 \times \frac{ - 8}{27} - \frac{4}{9 \times 3} - \bigg( \frac{ - 2}{3 \times 9} \bigg) + \frac{2}{27 }$

$\implies \: \frac{16}{27} - \frac{16}{27} - \frac{4}{27} + \frac{2}{27} + \frac{2}{27}$

$\implies \: \: \: \frac{16 - 16 - 4 + 2 + 2}{27}$

$\implies \: \: \frac{0}{27}$

$\boxed{\underline{ \implies \: 0}}$

Hence, required remainder is 0