Iknow how to subtract, the result is 3/x+1 but i don't know how to find the excluded numbers, kindly explain that

You cannot divide by zero, because that is undefined, it approaches infinity, which is not a real value.  Some would say that division by zero is "not allowed" :P  

So in this case the divisor cannot equal zero.

x^2+4x+3=0  There are many methods to factor, I'll complete the square, just for conciseness (and it is good to know how the quadratic formula is derived)

Divide by the leading coefficient, which is just 1 in this case...

Move the constant to the other side, subtract 3 from both sides

x^2+4x=-3  Halve the linear coefficient, square it, then add that value to both sides...(4/2)^2=2^2=4

x^2+4x+4=1  Now the left side is a perfect square...

(x+2)^2=1  Take the square root of both sides

x+2=±√1  subtract 2 from both sides

x=-2±√1

x=-2±1

x=-3 and -1

So if the equation equals zero when x=-3 and -1, it's factors are:

(x+3)(x+1)

So x cannot equal -3 or -1, otherwise your original equation would have division by zero.

The answer to your question is b’(16,14)