Could I factor w^2 + 4 w + 3?

Yes.

w^2 + 4 w + 3
=(w+3)(w+1)

w² + 4w + 3 = 0
w² + 4w = - 3
w² + 2w = - 3 + 2²
w² + 2w = - 3 + 4
(w + 2)² = 1
w + 2 = 1

1st factor:
= w + 2 - 1
= w + 1

2nd factor:
= w + 2 + 1
= w + 3

Answer: (w + 1)(w + 3) are the factors.

Proof:
= (w + 1)(w + 3)
= w² + 3w + w + 3
= w² + 4w + 3

if you mean factorize, yes you can easily.
= (w + 1)(w + 3)
if y= ax^2 + bx + c
the key is to look for two numbers that both
- add up to give 'b' and
- multiply to give 'c'.
it can sometimes take a long time to factorize, but eventually you'll get the hang of it and do it in no time :)

w^2 + 4w + 3
= w^2 + 3w + w + 3
= (w^2 + 3w) + (w + 3)
= w(w + 3) + 1(w + 3)
= (w + 3)(w + 1)

Yes. The above trinomial can be factored.

(w + 3) (w + 1)

You can check by multiplication using the FOIL Method:

w x w = w^2

w x 1 = 1w

3 x w = 3w

3 x 1 = 3

w^2 + 1w + 3w + 3

= w^2 + 4w + 3

w2+4w+3=w2+3w+w+1=w(w+3)+w+1

Yep,

w^2 + 4w + 3 = w^2 + 2*w*2 + 2^2 - 2^2 + 3 =
= (w^2 + 2*w*2 + 2^2) - 4 + 3 = (w + 2)^2 - 1 =
(w + 2 + 1)(w + 2 - 1) = (w + 3)(w + 1)

Yes. (w + 1)(w + 3)

(w+1)(w+3)

( w + 3 ) ( w + 1 )