Question

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

Answer 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)

Answer 2

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

✔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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