How do you factor #x^2 + 4x + 3#?

2 Answers
Jul 14, 2016


The factors are #(x+3)# and #(x+1)#


We are asked for factorise: #x^2+4x+3#

First notice that the function is a quadratic and so will have two factors. Since the coefficient of #x^2# is 1, the factors will be of the form: #(x+a)(x+b)# We will assume that a and b are integers.

Hence, we need to find a and b such that the product of the factors is equal to the given quadratic function .

Now consder the absoute value of constant term, 3. Since 3 is prime its only factors are 3 and 1. Since the constant term is positive, a and b can only be 3 and 1 or -3 and -1.

Finally observe that the coefficient of #x# is positive 4 and that the sum of 3 and 1 is positive 4. Thus, a and b must be 3 and 1 (or the other way around, but this makes no difference to our factorisation)

Hence we see that: #x^2+4x+3# = #(x+3)(x+1)#

Therefore the factors are #(x+3)# and #(x+1)#

Jul 14, 2016


(x + 1)(x + 3)


y = x^2 + 4x + 3
Since a - b + c = 0, use shortcut:
- One real root is (-1) and the factor is (x + 1)
- The other real root is (-c/a = - 3), and the factor is (x + 3)
y = (x + 1)(x + 3)