Question

Let f(x) be a Polynomial function Such that ,f(x²+1)=x⁴+5x²+3,then f(x²-1)=​

Answer 1

Given.

f(x²+1)=x⁴+5x²+3

Now it can be written as

f(x²+1)=[x⁴+5x²+6]-3

Now Factorise the given Polynomial.

f(x²+1)=[x⁴+3x²+2x²+6]-3

f(x²+1)=[x²(x²+3)+2(x²+3)]-3

f(x²+1)=[(x²+2)(x²+3)]-3

Now It can Be written as

f(x²+1)=[(x²+1+1)(x²+1+2)]-3

Now We have to Find X²-1

So Subsitute it on place of x²+1 in Polynomial.

so

f(x²-1)=[(x²-1+1)(x²-1+2)]-3

f(x²-1)=[x²(x²+1)]-3

f(x²-1)=[x⁴+x²-3]

f(x²-1)=[x⁴+x²-3]

Answer 2

Step-by-step explanation:

f(x²+1)=x⁴+5x²+3

f(x²+1)=(x⁴+5x²+6)-3

=x²(x²+3)+2(x²+3)

f(x²+1) =(x²+3)(x²+2)

f(x²-1)=(x²-1+2)(x²-1+1)

f(x²-1) =(x²-1)x²

this is your answer ❣️❣️✌️✌️