Question

Find all polynomials P(x) such that (p(x))^2 =1+ x.P(x+1) for all real numbers x.

Answer 1

Answer:

p(x) = x + 1

Step-by-step explanation:

Let the degree of the polynomial p(x) be 'n'

Given p²(x) = 1 + x.p(x+1),

Now we should note that if degree of polynomial p(x) is 'n', then the

degree of polynomial p²(x)  will be '2n'

degree of polynomial p(x+1) will be 'n' itself,

hence L.H.S is of degree '2n'

R.H.S is of degree 1+n

=> 2n = n + 1

=> n = 1.

Hence p(x) is polynomial of degree 1.

Let us assume p(x) be ax +b.

Now using the given equation

(ax+b)² = 1 + x(a(x+1) + b)

=> a²x² + b² + 2abx = 1 + ax² + ax + bx

=>(a²-a)x² + (2ab-b - a)x + b²-1 = 0

Since a general polynomial to be zero , all its coefficients should be 0

=> a² = a

=> a = 0 or a = 1

a cannot be 0, since p(x) is of degree 1, hence a = 1

and

2ab - b - a = 0

=>b = 1

Hence , there is only one polynomial p(x) = x + 1.

Hope, it helped !