00:01
Hi, here for the given question we need to show that the division algorithm does not hold for zx.
00:20
So here in our case we will prove this step.
00:24
Let us assume contrary.
00:26
So here let us suppose that it holds for z of x.
00:38
So here in our case we can assume that there exists a function f of x and g of x in zx which are non -zero polynomial and g of x is non -constant polynomial.
01:05
Then there exists a unique polynomial q of x and r of x which belongs to zx such that the value of f of x equals to g of x multiplied with q of x plus r of x.
01:32
So here in our case we can say that the degree of r of x is less than degree of g of x or r of x is a zero polynomial.
01:56
Now here we can also check this by taking two polynomial.
02:01
Let f of x equals to 3x square plus 5 and g of x equals to 2x plus 5.
02:11
Now here we can observe that if degree of g of x is equal to 1 then r of x is constant because the degree will be less than 1 which is 0.
02:28
So it will be constant.
02:29
So r of x equals to c which belongs to z.
02:34
Therefore here we can say that we have 3x square plus 5 equals to 2x plus 5 multiplied with qx plus c.
02:44
So here further we have 3x square plus 5 plus c equals to 2x plus 5 multiplied with qx...