The question of class 10th of exercise 2.3 of question number 5. How can i solve all the parts of this question in easy way. This is very important question of class 10th Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0

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# Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0 Q.5

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According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

Dividend = Divisor × Quotient + Remainder

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Now let us proof the three given cases as per division algorithm by taking examples for each.

(i) deg p(x) = deg q(x)

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us take an example, p(x) = 3x

^{2}+3x+3 is a polynomial to be divided by g(x) = 3.So, (3x

^{2}+3x+3)/3 = x^{2}+x+1 = q(x)Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).

Hence, division algorithm is satisfied here.

(ii) deg q(x) = deg r(x)

Let us take an example, p(x) = x

^{2 }+ 3 is a polynomial to be divided by g(x) = x – 1.So, x

^{2 }+ 3 = (x – 1)×(x) + (x + 3)Hence, quotient q(x) = x

Also, remainder r(x) = x + 3

Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).

Hence, division algorithm is satisfied here.

(iii) deg r(x) = 0

The degree of remainder is 0 only when the remainder left after division algorithm is constant.

Let us take an example, p(x) = x

^{2 }+ 1 is a polynomial to be divided by g(x) = x.So, x

^{2 }+ 1 = (x)×(x) + 1Hence, quotient q(x) = x

And, remainder r(x) = 1

Clearly, the degree of remainder here is 0.

Hence, division algorithm is satisfied here.