Factors are (x + 1)(x + 3) so x = -1 or -3.
* * *
The factors given above are correct.
But the solution offered above is correct, only if
x2 + 4x + 3 = 0.
If the given quadratic function is equated to anything other than zero, then the solution will also be something else.
Example:
x2 + 4x + 3 = 15 gives
x2 + 4x - 12 = 0, whence,
(x - 2 )(x + 6) = 0; and
it is evident that x = 2 or -6 is the desired solution.
In other words, if you substitute either 2 or -6 for x in the equation
x2 + 4x + 3 = 15,
then, that equation becomes a true statement. Test it, and see!
On the other hand, x = -1 or -3 makes the equation
x2 + 4x + 3 = 0
a true statement.
The moral to remember is that expressions, such as the asker gave above, have factors; but only equations have solutions.
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2 + 4x + 3 = (x + 1)(x + 3)
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
You can work this out with long division, by checking to see if (x2 - 1) is a factor of (2x4 + 4x3 - x2 + 4x - 3). It is. Unfortunately, the WikiAnswers system is somewhat limited in depicting things such as long division, so we won't be able to represent it here. In short though, (2x4 + 4x3 - x2 + 4x - 3) / (x2 + 1) is equal to 2x2 + 4x - 3. which means that: (x2 + 1) / (2x4 + 4x3 - x2 + 4x - 3) = (x2 + 1) / (x2 + 1)(2x2 + 4x - 3) = 1 / (2x2 + 4x - 3)
y = 4x-3
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2 - 4x + 3 is already in standard form.
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
x2 + 4x + 3 =(x + 3) (x + 1)
(x-1)(x-3)