Help me factor x^2 -2x - 4 = 0 and 3x^2 + x + 3 = 0.

No factor

you can tell there are no rational roots because the discriminant is negative. No real roots, no factoring.

You can factor over the irrationals or complex, though, getting

(x-1-āˆš5)(x-1+āˆš5)
and
1/12 (6x+1-iāˆš35)(x+1+iāˆš35)

To factor quadratic equations like these, you need to find two binomials that multiply together to give you the original equation. Let's go step by step.

1. Factor x^2 - 2x - 4 = 0:
To factor this equation, we need to find two binomials in the form (x + a)(x + b) where a and b are constants. Since the coefficient of x^2 is 1, we know that (x + a)(x + b) will have the form x^2 + (a + b)x + ab.

If we compare the given equation x^2 - 2x - 4 to x^2 + (a + b)x + ab, we can see that a + b = -2 and ab = -4.

To find a and b, we need to find two numbers whose sum is -2 and whose product is -4. After some trial and error, we can see that a = -4 and b = 1 satisfy these conditions.

Therefore, the factored form of x^2 - 2x - 4 = 0 is (x - 4)(x + 1) = 0.

2. Factor 3x^2 + x + 3 = 0:
Similarly, we need to find two binomials in the form (x + a)(x + b) that multiply together to give 3x^2 + x + 3.

Again, we need to find two numbers whose sum is 1 (the coefficient of x) and whose product is the product of the coefficient of x^2 and the constant term. In this case, that is 3*3 = 9.

By trial and error, we can see that a = -3 and b = -1 satisfy these conditions.

Therefore, the factored form of 3x^2 + x + 3 = 0 is (x - 3)(x - 1) = 0.

So, the factored forms of x^2 - 2x - 4 = 0 and 3x^2 + x + 3 = 0 are (x - 4)(x + 1) = 0 and (x - 3)(x - 1) = 0, respectively.