can you show the steps

x^{2} + 3x + 5

For equations of the form:

ax^{2} + bx + c

where a = 1 it will factor to

(x + c_{1})(x + c_{2})

If you multiply that out you'll end up with

x^{2} + c_{1}x + c_{2}x + c_{1}*c_{2} = x^{2} + (c_{1} + c_{2})x + c_{1}*c_{2}

This means that

b = c_{1}+c_{2} and is the sum of the factors of c

c = c_{1}*c_{2} and is the product of the factors of c

Looking back at your equation

a = 1, b = 3, and c = 5

The only factors of a prime number are 1 and the number so c = 1*5

In order to factor it b has to equal the sum of those two numbers so it would be 6x not 3x.

This is why your equation cannot be factored in the standard "normal" way.

Using the quadratic formula:

x = [-b ± √(b^{2} - 4ac)]/2a

b^{2} - 4ac = 9 - 20 = -11

so you have complex roots and

x = (-3+ i√11)/2

x = (-3 - i√11)/2

In factored form that would look like

(x + 3/2 - i√11/2)(x + 3/2 + i√11/2)

Multiplying that out to check you'll find

x^{2} + (3/2)x +
(i√11/2)x + (3/2)x + 9/4 + (3i√11)/4 - (i√11/2)x - (3i√11)/4 - i^{2}(√11)^{2}/4

Since i^{2} = -1 and (√11)^{2} = 11, the term on the end simplifies to 11/4

and your equation is

x^{2} + (3/2)x + (3/2)x + 9/4 + 11/4

x^{2} + 3x + 5