First, get rid of that pesky #1/2#. Fractions are usually difficult to work with, but in this case we can factor out the #1/2# and not have to deal with them at all:

#f(x)=1/2(x^2+5x-3)#

Now we can ignore the #1/2# and focus on the good stuff:

#x^2+5x-3#

We must ask ourselves, "Is there a number pair that multiplies to #-3# and adds to #5#?" We will quickly discover that no such pair exists (no rational numbers meet these criteria), and indeed, we must use the quadratic formula. This is due to fact that the determinant #b^2-4ac# is not a perfect square.

This equation will have roots at:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#=(-5+-sqrt(5^2-4(1)(-3)))/(2(1))#

#=(-5+-sqrt(25+12))/(2)#

#->x=(-5+-sqrt(37))/(2)#

If #x=a# is a root of a polynomial, then #x-a=0#, and #(x-a)# is a factor of the polynomial. So if we want to factor #x^2+5x-3#, we'll need the roots to look something like #x-"blah blah"=0#.

Let's get to it. We have:

#x=(-5+-sqrt(37))/(2)#

Doing some algebra, we can simplify this to:

#2x+5=+-sqrt(37)#

We have two cases: #sqrt(37)# and #-sqrt(37)#. We'll work with them both:

#2x+5=sqrt(37)->2x+5-sqrt(37)=0#

#2x+5=-sqrt(37)->2x+5+sqrt(37)=0#

So the two factors of our quadratic are #2x+5-sqrt(37)=0# and #2x+5+sqrt(37)=0#, and that means we're done.

But note that multiplication of these monomials leads to #4x^2# not #1/2x^2#, so we should divide the product by #8#. Hence,

#f(x)=1/8(2x+5-sqrt(37))(2x+5+sqrt(37))#