### Chapter 2: Polynomials

Q
##### Polynomials CBSE NCERT Solutions

Question:

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

Solutions: 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 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)

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

Let us take an example, 3x2 + 3x + 3 is a polynomial to be divided by 3.

So, = x2 + x + 1 = q(x)

Thus, we can see, the degree of the quotient is equal to the degree of dividend.

Hence, division algorithm is satisfied.

(ii)  deg q(x) = deg r(x)

Let us take an example, p(x) = 2x3 - 2x2 + 2x + 3 is a polynomial to be divided by g(x) = 2x2 - 1.

So, $\frac{2x^{3}-2^x{2}+2x+3}{2x^{2}-1}$  =  x-1 = q(x)

Also, remainder, r(x) = 3x+2

Thus, you can see, the degree of the quotient is equal to the degree of the remainder.

Hence, the division algorithm is satisfied.

(iii) deg r(x) = 0

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

Let us take an example, p(x) = x+ 1 is a polynomial to be divided by g(x)= x.

So, q(x) = x and r(x) = 1

The degree of remainder here is 0.

Hence, the division algorithm is satisfied.

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