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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**

Answer:

**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, 3x ^{2 }+ 3x + 3 is a polynomial to be divided by 3.**

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

**So, ** **= 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 ^{2 }+ 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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