How would you factor 3x^2 - 5x - 7?

Question

and could you please tell me how you go the answer too? I'm still so confused on how to factor =(

Answer 1

3x^2 - 5x - 7 does not have integer factors. It can be factored by completing the square. It will look a little messy--
First, "factor out" the 3 to get x^2 alone:
3(x^2 - (5/3)x - 7/3) =
Now, complete the square:
3(x^2 - (5/3)x + (5/6)^2 - (5/6)^2 - 7/3) =
Resolve the "left-over" numbers:
3(x - 5/6)^2 - (25/36) - 7/3) =
3(x - 5/6)^2 - 25/36 - 84/36) =
3(x - 5/6)^2 - 109/36) =
Now, factoring the "difference of squares" (sadly, 109 is prime)
3(x - 5/6 - (1/6)√109)(x - 5/6 + (1/6)√109)

Answer 2

In general, if xo is one solution of the equation f(x) = 0, where f(x) is one polynomial then f(x) can be factorized as:
f(x) = (x-xo)*g(x)
You can find g(x) by dividing f(x)/(x-xo)
Do the same with g(x) you can step by step to factorize f(x) into:
f(x) = (x-xo)*(x-x1)*(x-x2).....
In your case, f(x) = 3x^2 - 5x -7 = 0 is one quadratic equation with two solutions, say x1, x2 (I hope that you can calculate) so:
f(x) = A*(x-x1)*(x-x2)

Answer 3

I have tried to factor this and it does not factor

If you needed to find out what x is you would have to use the quadratic