Math, asked by Anonymous, 9 months ago

Please answer as soon as possible. Please don't copy and paste for other sites bcz I already seen their answers.

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Answered by Anonymous
1

Let p(x) and g(x) be two polynomials

If g(x) is any polynomial then it can divide p(x) by q(x) where 0<q(x) and may get a remainder say r(x).

If g(x) perfectly divides p(x) by q(x), then r(x)=0.

It is obvious that deg r(x)<deg g(x).

∴ we can find polynomial q(x) and r(x) such that

p(x)=q(x)q(x)+r(x), where r(x)=0 or deg r(x)<deg g(x)

Answered by Anonymous
85

\huge{\bold{\star{\fcolorbox{black}{red}{Solution...}}}}

✔1️⃣. deg p(x) = deg q(x) ☃️

We know the formula:

\sf\large\pink{Divisor × quotient + remainder}

\rm\large{\implies\ p(x) = g(x) × q(x) + r(x)}

☆So, here the degree of quotient will be equal to degree of Divident when the divisor is constant.

□Let is assume the division of \rm{4^{2} by 2}

Here, \rm{p(x) = 4x^{2}}

\rm{q(x) = 2x^{2} and r(x) = 0}

Degree of p(x) and q(x) is the same, i,e. 2.

Checking for division algorithm,

\rm\large{p(x) = g(x) × q(x) + r(x)}

\rm\large{4x^{2} = 2(2x^{2}) }

Hence the division algorithm is satisfied.

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2️⃣. deg g(x) = deg r(x)

Let us assume the division of \rm\large{x^{3} + x\: by \:x^{2}}

Here,

\rm\large{p(x) = x^{3} + x, g(x) = x^{2}, q(x) = x \:and\: r(x) \:=\: x}

Degree of q(x) and r(x) is the same I.e. 1.

Checking for division algorithm,

\rm\large{p(x) = g(x) × q(x) + r(x) }

\rm\large{x^{3} + x = x^{2} × x + x}

\rm\large{x^{3} + x = x^{3} + x }

Hence, the division algorithm is satisfied!

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3️⃣. deg r(x) = 0

Degree of remainder will ne zero when remainder comes to a constant.

Let us assume the division of \rm\large{x^{4} +1 by x^{2}}

Here,

\rm\large{p(x) = x^{4} + 1}

\rm\large{g(x) = x^{3}}

\rm\large{q(x) = x \:and\: r(x) = 1}

\tt\large\red{Degree \:of\: r(x) \:is \:zero}

Checking for division algorithm,.

\rm\large{p(x) = g(x) × q(x) + r(x)}

\rm\large{x^{4} + 1 = x^{3} × x + 1}

\rm\large{x^{4} +1 = x^{4} + 1 }

Hence, the division algorithm is satisfied.☃️

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