Linear Function

why is there no pair of linear functions that can multiply together to result f(x)= x^2+1

There is a pair of linear functions: 

$\begin{array}{rcll} x^2+1 &=& (x-i)\cdot (x+i) \qquad & | \qquad i \text{ is the imaginary number} \end{array}$

$\begin{array}{rcll} x^2+1 &=& (x-i)\cdot (x+i) \\ &=& x^2 - i^2 \qquad & | \qquad i^2 = -1 \\ &=& x^2 - (-1) \\ &=& x^2 +1 \end{array}$

Mmm thanks  Heureka,

most kids here have not ever dealt with imaginary numbers so I think the question refers only to real numbers.

Lets see.

why is there no pair of linear functions that can multiply together to result f(x)= x^2+1

I am just thinking here....

Let the 2 linear functions be  f(x)=ax+k      and     g(x)=mx+b    where a,k,m, and b are all real numbers.

$(ax+k)(mx+b) = amx^2 + abx  + kmx  + kx\\ (ax+k)(mx+b) = amx^2 + (ab+km)x  + kb$

Now we want this to equal x^2+1

$amx^2 + (ab+km)x  + kb =x^2+1\\ $

Equating coefficients

am=1     a=1/m

bk=1      b=1/k

ab+km=0    1/(mk) +km=0          1/(mk) = -mk      

This is impossible becaus one side will be negative and the other will be positive.

Neither m nor k can be zero becasue you cannot divide by 0.

Hence 2 linear functions cannot multiply to give   x^2+1   

(not in the real number system anyway )     :))

While the outline of what I have done should be correct my layout is very poor.

Perhaps someone can present it better ??

Let's suppose that there could be.....call the functions

f(x)  = ax + b      and  g(x)  = cx + d

So ......   a and c must be reciprocals and so must b and d....so we have

(ax + b) ((1/a)x + (1/b)  = 

x^2 + (b/a)x + (a/b)x + 1      

And this = x^2 + 1

Which implies that

[ (b/a) + (a/b)]    = 0

Which implies that

[b^2 + a^2] / ab   = 0

Which implies that

b^2 + a^2   = 0

But this is only possible if a and b  = 0

But....if a and b = 0 

Then f(x)  =  0x +  0   = 0

And   g(x)   = (1/0)x + (1/0)

Thus :   f(x) * g(x)   = 0 times an undefined expression   ....which could never be any function!!!