Solve by factorizing
6x^2 -x -12=0
Apparently it equals to zero, but im not sure how they got it, so could someone help please????
Do you know how to factorise 6x^2 - x - 12? See Wolfram|Alpha
Equals to zero. That's what happens when all terms are one side to the extent of my knowledge since this one definitely does factor.
$\displaystyle (3x+4) (2x-3)$
$\displaystyle 6x^2 -9x+ 8x -12. $
$\displaystyle 6x^2 - x - 12. $ Checks out.
You just have to realize what'll work and what won't.
Try plugging in numbers that you think will get you 6x^2 but at the same time think of two numbers that will get you the other two numbers. Look at the factors of 12. 1, 2, 3, 4, 6, 12. Try them. 1 x 12, 2 x 6, 3 x 4 are the pairs.There are two ways to get six. 1, 2, 3, 6. 2 x 3 or 1 x 6. And it should become pretty intuitive to you once you get how factoring works.
fvaras89: "Solve by factorizing, 6x^2 -x -12=0. Apparently it equals to zero, but im not sure how they got it, so could someone help please?"
6x^2 -x -12 = (2x-3) (3x+4), but why?
without the aid of the computer or calculator, you are left to solve by hand,
6x^2 -x -12 = 0, can be factored in this form, (ax + b)(cx + d)
factor of 6: 6 & 1, 3 & 2, 2 & 3.
factor of -12: -12 & 1, -1 & 12, -6 & 2, -2 & 6, -4 & 3, -3 & 4
Now analyze what pairing will generate 6x^2 -x -12.
We choose 6: 2 & 3,
and for -12: -3 & 4.
Now we have,
A) (2x + 4)(3x - 3) or
B) (2x - 3)(3x + 4).
Just check the MIDDLE term for
A) middle term = 2x(-3) + 4(3x) = -6x + 12x = 6x, wrong result, next
B) middle term = 2x(4) + (-3)(3x) = 8x - 9x = -x, YES!
Now, our factorization is, 6x^2 -x -12 = (2x-3) (3x+4).
ok - - - - -
You have provided no explanation of what you are trying to do or why. Maths is not some magical performance where one waves ones hands does this does that and the answer magically appears.
So in this case you explain that the rational roots theorems tells you that if this quadratic has any rational roots they are or the form of the ratio of a factor of the constant term to a factor of the coefficient of the $\displaystyle x^2$ term.
CB