Question

yrr jisko aata h wohi krna plz glt answer Mt krna ​

Answer 1

Step-by-step explanation:

SEE THE ATTACHMENT........

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Answer 2

$\underline{\bigstar\:\textsf{According \: to \: given \: in \: question:}}$

$\normalsize\:\bullet\:\sf\ According \: to \: the \: division \: algorithm, \: if \: P(x) \: and \: g(x) \\ \normalsize\sf\ are \: two \: polynomial \: with \: g(x) = 0, \: then \: we \: can \: find \\ \normalsize\sf\ P(x) \: and \: r(x) \: such \: that \: P(x) = g(x) \times\ q(x) + r(x) \\ \normalsize\sf\ [where \: r(x) = 0].$

$\normalsize\sf\:\bullet\ Degree \: of \: polynomial \: is \: the \: highest \: power \\ \normalsize\sf\ of \: the \: variable \: in \: the \: polynomial.$

$\rule{100}1$

AnswEr 1 :

$\:\bullet\:\sf\ Degree \: of \: quotient \: will \: be \: equal \: to \: Degree \\ \normalsize\sf\ of \: dividend \: when \: divisor \: is \: constant \\ \normalsize\sf\ [i.e. - when \: p(x) \: is \: divided \: by \: constant].$

$\normalsize\sf\ Let \: us \: assume \: division \: of \: 6x^2 + 2x + 2 \: by \: 2$

$\normalsize\dashrightarrow\sf\ P(x) = 6x^2 + 2x + 2 \\ \\ \normalsize\dashrightarrow\sf\ g(x) = 2 \\ \\ \normalsize\dashrightarrow\sf\ q(x) = 3x^2 + x + 1 \\ \\ \normalsize\dashrightarrow\sf\ r(x) = 0$

$\normalsize\sf\ Here, \: deg \: p(x) = deg \: q(x) \: [i.e. 2]$

$\scriptsize\sf{\: \: \: \: \: \: \: \: \: \dag\ \: Checking \: for \: division \: algorithm}$

$\normalsize\dashrightarrow\sf\ P(x) = g(x) \times\ q(x) + r(x)$

$\normalsize\dashrightarrow\sf\ 6x^2 + 2x + 2 = 2[ 3x^2 + x + 1 + 0 ]$

$\normalsize\dashrightarrow\sf\ 6x^2 + 2x + 2 = 2[ 3x^2 + x + 1]$

$\normalsize\dashrightarrow\sf\ 6x^2 + 2x + 2 = 6x^2 + 2x + 2$

$\normalsize\dashrightarrow\sf\ Hence, \: algorithm \: verified \: !!$

$\rule{100}1$

AnswEr 2 :

$\normalsize\sf\ Let \: us \: assume \: the \: division \: of \: x^3 + x \: by \: x^2$

$\normalsize\dashrightarrow\sf\ P(x) = x^3 + x \\ \\ \normalsize\dashrightarrow\sf\ g(x) = x^2 \\ \\ \normalsize\dashrightarrow\sf\ q(x) = x \\ \\ \normalsize\dashrightarrow\sf\ r(x) = x$

$\normalsize\sf\ Here, \: deg \: q(x) = deg \: r(x) \: [i.e. \: 1]$

$\scriptsize\sf{\: \: \: \: \: \: \: \: \: \dag\ \: Checking \: for \: division \: algorithm}$

$\normalsize\dashrightarrow\sf\ P(x) = g(x) \times\ q(x) + r(x)$

$\normalsize\dashrightarrow\sf\ x^3 + x = x^2[ x ] + x$

$\normalsize\dashrightarrow\sf\ x^3 + x = x^3 + x$

$\normalsize\dashrightarrow\sf\ Hence, \: algorithm \: verified \: !!$

$\rule{100}1$

AnswEr 3 :

$\normalsize\sf\ Degree \: of \: remainder \: will \: be \: 0 \: when \\ \normalsize\sf\ remainder \: comes\: to \: a \: constant$

$\normalsize\sf\ Let \: us \: assume \: the \: division \: of \: x^3 + 1 \: by \: x^2$

$\normalsize\dashrightarrow\sf\ P(x) = x^3 + 1 \\ \\ \normalsize\dashrightarrow\sf\ g(x) = x^2 \\ \\ \normalsize\dashrightarrow\sf\ q(x) = x \\ \\ \normalsize\dashrightarrow\sf\ r(x) = 1$

$\normalsize\sf\ Here, \: deg \: r(x) = 0$

$\scriptsize\sf{\: \: \: \: \: \: \: \: \: \dag\ \: Checking \: for \: division \: algorithm}$

$\normalsize\dashrightarrow\sf\ P(x) = g(x) \times\ q(x) + r(x)$

$\normalsize\dashrightarrow\sf\ x^3 + 1 = x^2[ x ] + 1$

$\normalsize\dashrightarrow\sf\ x^3 + 1 = x^3 + 1$

$\normalsize\dashrightarrow\sf\ Hence, \: algorithm \: verified \: !!$