How do you simplify the expression #(x^2 + 4x - 3) /( x^3 + 9x^2 + 8x)#?

To investigate this, factorise the numerator and denominator and compare to see if any factors cancel out. It is often useful to compare coefficients to attempt to 'guess' the factors, but it is always possible to reliably find them using the quadratic formula.

Recall the quadratic formula, derivable by the method of "completing the square":
For equation #ax^2+bx+c=0#, #x=(-b+-sqrt(b^2-4ac))/(2a)#

Numerator
Set this equal to 0 and solve as a quadratic equation:
#x^2+4x-3=0#

This gives us #x=(-4+-sqrt(16+12))/2=(-4+-sqrt(28))/2=(-4+-2sqrt(7))/2#
So #x=-2+-sqrt(7)# and #x^2+4x-3# factorises as #(x+2-sqrt(7))(x+2+sqrt(7))#. This is not usually considered a simplification and so the numerator here would be expressed in its simplest form as given in the question.

Denominator
Note that the factor #x# can be removed to make the expression #x(x^2+9x+8)#.
Solve the quadratic factor as before:
#x^2+9x+8=0#
#x=(-9+-sqrt(81-32))/2=(-9+-sqrt(49))/2=(-9+-7)/2=-1,-8#

So #x^2+9x+8=(x+1)(x+8)#

Thus the denominator factorises as
#x(x+1)(x+8)#

Simplified form

It is arguable whether it is simpler to leave the expression without brackets in the original form given or to factorise it as much as possible while retaining integer coefficients. The factorised form of the original expression is:
#(x^2+4x-3)/(x(x+1)(x+8)#

I wonder if there is a typo in the question - if the numerator were instead #x^2+4x+3#, then this factorises as #(x+1)(x+3)#, giving us a common factor of #x+1# to cancel out, leaving
#(x+3)/(x(x+8))#