Question

Give an example of polynomials f(x),g(x),q(x) and r(x) satisfying f(x)=g(x),q(x)+ r(x), where degree r(x)=0

Answer 1

SOLUTION :

Let f(x) = 2x⁴ + 8x³ + 6x² +4x +12

q(x) = 2

g(x) = x⁴ + 4x³ + 3x² + 2x + 1

r(x) = 10

Here, degree r(x) = 0

By using division algorithm ,

Dividend =  Divisor ×  Quotient + Remainder

f(x) = q(x)  × g(x) + r(x)

= 2( x⁴ + 4x³ + 3x² + 2x + 1) + 10

= 2x⁴ + 8x³ + 6x² +4x + 2 +10

= 2x⁴ + 8x³ + 6x² +4x +12

Hence, above values of  f(x) ,q(x) , g(x) and r(x) satisfy the division algorithm.

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Answer 2

i) Let us assume the division of 6x2 + 2x + 2 by 2
Here, p(x) = 6x2 + 2x + 2
g(x) = 2
q(x) = 3x2 + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x2 + 2x + 2 = 2x (3x2 + x + 1)
Hence, division algorithm is satisfied.


(ii) Let us assume the division of x3+ x by x2,
Here, p(x) = x3 + x
g(x) = x2
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x3 + x = (x2 ) × x + x
x3 + x = x3 + x
Thus, the division algorithm is satisfied.

(iii) Let us assume the division of x3+ 1 by x2.
Here, p(x) = x3 + 1
g(x) = x2
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x3 + 1 = (x2 ) × x + 1
x3 + 1 = x3 + 1
Thus, the division algorithm is satisfied.