Question

$if \: the \: polynomial \: x {}^{4} - 6 {x}^{3} + 16 {x}^{2} - 25 {x}^{} + 10 \: i s \: divided \: by \: another \: polynomial \: {x}^{2} - 2 {x}^{} + k \: the \: remainder \: comes \: out \: to \: be \: x + a \: find \: k \: and \: a \:$
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Answer 1

${ \huge{ \pink{ \boxed{ \mathcal{ \purple{ÃÑẞWĒR}}}}}}$

Step-by-step explanation:

$\pink{We \: know \: that,}$

$Dividend = Divisor × Quotient + Remainder$

$⇒ Dividend – Remainder  =  Divisor × Quotient$

$Dividend \: – \: Remainder \: is \: always \: divisible \: by \: the \: divisor.$

$\boxed{Now, \: it \: is \: given \: that \:  f(x)   \: when \: divided \: by   \: {x}^{2}  – 2x + k  \:  leaves \:  (x + a) \: as \: remainder.}$

$So, \:  for   \: f(x) \: to \: be \: completely \: divisible \: by   {x}^{2}  – 2x + k,  remainder \: must \: be \: equal \: to \: zero$

$⇒  (–10 + 2k)x + (10 – a – 8k +  {k}^{2} ) = 0$

$⇒  –10 + 2k = 0  and  10 – a – 8k +  {k}^{2}  =  0$

$⇒  k = 5  and  10 – a – \: 8 (5) + {5}^{2}  = 0$

$⇒  k = 5  and   – a – 5 = 0$

$⇒  k = 5  and   a  =  –5$

${ \bf{ \blue{i \: hope \: it \: helpfull \: for \: you}}}$

Answer 2

Answer:

Step-by-step explanation:

U will get in the pic

I hope this helps