Question

Find the remainder,when
${x}^{3} + 3 {x}^{2} + 3x + 1 \: is \: divided \: by \: (5 + 2x)$
​

Answer 1

the question is wrong because 2 X + 5 is not divided by x cube + 3 X square + 3 X + 1

Answer 2

Q. Find the remainder,when

Find the remainder,when ${x}^{3} + 3 {x}^{2} + 3x + 1 \: is \: divided \: by \: (5 + 2x)$

Explanation:-

$we \: have \: p(x) = {x}^{3} + 3 {x}^{2} + 3x + 1$

$and \: divisior \: is \: (5 + 2x)$

By remainder theorm,the required remainder is :-

$5 + 2x = 0$

$x = \frac{ - 5}{2}$

$p = ( \frac{ - 5}{2} )$

By putting the value in X

$p( \frac{ - 5}{2} ) = ( { \frac{ - 5}{2}) }^{3} + 3 \times ( { \frac{ - 5}{2} )}^{2} + 3 \times ( \frac{ - 5}{2} ) + 1$

$= \frac{ - 128}{5} + 3 \times \frac{25}{4} - \frac{15}{2} + 1$

$= \frac{ - 125}{8} + \frac{75}{4} - \frac{15}{2} + 1$

$= \frac{ - 125 + 150 - 60 + 8}{8}$

$= \frac{ - 27}{8}$

$hence \: the \: remainder \: is \: = \frac{ - 27}{8}$