Q: 5 Give examples of polynomial 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 Give examples of polynomial 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 Give examples of polynomial 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 Give examples of polynomial 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

Asked by Ashish Parashar | 21st Sep, 2013, 09:03: AM

Expert Answer:

According to the division algorithm, if p(x) and g(x) are two polynomials with g(x)  0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) x q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x).

(i)    Degree of quotient will be equal to degree of dividend when divisor is constant.
Let us consider the division of  by 3.
Here, p(x) =   and g(x) = 3
q(x) =   and r(x) = 0
Here, degree of p(x) and q(x) is the same which is 2.


Checking:
p(x) = g(x) x q(x) + r(x)

 

Thus, the division algorithm is satisfied.

(ii)    Let us consider the division of 2x4 + 2x by 2x3,
Here, p(x) = 2x4 + 2x and g(x) = 2x3
q(x) = x and r(x) = 2x
Clearly, the degree of q(x) and r(x) is the same which is 1.

Checking,
p(x) = g(x) x q(x) + r(x)
2x4 + 2x =  (2x3 ) x x  + 2x
2x4 + 2x = 2x4 + 2x
Thus, the division algorithm is satisfied.

(iii)    Degree of remainder will be 0 when remainder obtained on division is a constant.
Let us consider the division of 10x3 + 3 by 5x2.
Here, p(x) = 10x3 + 3 and g(x) = 5x2
q(x) = 2x and r(x) = 3
Clearly, the degree of r(x) is 0.

Checking:
p(x) = g(x) x q(x) + r(x)
10x3 + 3 = (5x2 ) x 2x  +  3
10x3 + 3 = 10x3 + 3
Thus, the division algorithm is satisfied.
 
Note: We have provided one example, for each case. You can find more on similar lines.

Answered by  | 22nd Sep, 2013, 03:51: PM

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