Prove that p(x)=g(x)×q(x)+r(x).
Answers
Answer:
ANSWER
(i) deg p (x) = deg q (x)
We know the formula,
Dividend = Divisor x quotient + Remainder
p(x)=g(x)×q(x)+r(x)
So here the degree of quotient will be equal to degree of dividend when the divisor is constant.
Let us assume the division of 4 x
2
by 2.
Here, p(x)=4 x
2
g(x)=2
q(x)= 2 x
2
and r(x)=0
Degree of p(x) and q(x) is the same i.e., 2.
Checking for division algorithm,
p(x)=g(x)×q(x)+r(x)
4 x
2
=2 ( 2 x
2
)
Hence, the division algorithm is satisfied.
(ii) deg q (x) = deg r (x)
Let us assume the division of x
3
+x by x
2
,
Here, p(x) = x
3
+x, g(x) = x
2
, q(x) = x and r(x) = x
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)
x
3
+x=x
2
×x+x
x
3
+x=x
3
+x
Hence, the division algorithm is satisfied.
(iii) deg r (x) = 0
Degree of remainder will be 0 when remainder comes to a constant.
Let us assume the division of x
4
+1 by x
3
Here, p(x) = x
4
+1
g(x) = x
3
q(x)=x and r(x)=1
Degree of r(x) is 0.
Checking for division algorithm,
p(x)=g(x)×q(x)+r(x)
x
4
+1=x
3
×x+1
x
4
+1=x
4
+1
Hence, the division algorithm is satisfied.
hope it will help you...