I think the attachment sums it up...Where did I go wrong?
why do you complicate...you can use determinant
2x(squared)+x-6.......ax(squared)+b+c..a=2, b=1, c=-6
d(stands for determinant cause i can't write its symbol)
d=b(squared)-4ac
d=1(squared)-4(2)(-6)
d=1+48
d=49
square root(d)=7
x1=(-b-(square root)d)/2a=(-1-7)/4=-8/4=-2
x2=(-b+(squared root)d/2a=(-1+7)/4=6/4=3/2
the factorization=(x-x1)(x-x2)=(x+2)(x-3/2)...you can change 3/2 and you will get
(x+2)(2x-3)
If you know how to factor when the coefficient of x^2 is one, here is the way to transform a quadratic equation with leading term not equal to one to quadratic equation with coefficient 1.
$\displaystyle 2x^2+x-6=0$
Multiply by coefficient of x^2 (which is 2)
$\displaystyle 4x^2+2x-6=0$
$\displaystyle (2x)^2+2x-6=0$
Let $\displaystyle a=2x$, then the above is the same as
$\displaystyle a^2+a-6=0$
Now you should be able to factor it!!
$\displaystyle (a+3)(a-2)=0$
$\displaystyle a=-3 or a=2$
$\displaystyle 2x=-3 ,2x=2$
$\displaystyle x=-3/2 , x=1$
My equation: 2x^2 + 4x - 6
Having swapped the 2 and 3: 2x + x - 6
But I still don't see where I'm wrong. I mean, I should be able to say "- 2x + 3x = b", where b is the "+x" in the middle of the equation, and I can. I don't see how I would tell that I'm wrong, and how to avoid the mistake in the future, short of expanding it again.
HelloMy equation: 2x^2 + 4x - 6
Having swapped the 2 and 3: 2x^2 + x - 6
But I still don't see where I'm wrong. I mean, I should be able to say "- 2x + 3x = b", where b is the "+x" in the middle of the equation, and I can. I don't see how I would tell that I'm wrong, and how to avoid the mistake in the future, short of expanding it again.
It's good to see that you are trying to figure it out.
When the coefficient of $\displaystyle x^2$ is 1 then you are correct, but you have $\displaystyle 2x^2$, so expanding.
(2x - 2)(x + 3)
$\displaystyle 2x^2 + 6x -2x -6$
b = 2*3 - 2 = 4
Whereas
(2x - 3)(x + 2)
$\displaystyle 2x^2 + 4x -3x -6$
b = 2*2 - 3 = 1
When the coefficient of $\displaystyle x^2$ is greater than 1 then one of the pair of factors of the constant term, c, will be multiplied by that coefficient. You have to consider that when finding which pair of factors, have a sum or difference equal to b.
Is that any clearer?
Try factoring some of these, maybe, to get the idea.
$\displaystyle 2x^2 + 5x + 2$
$\displaystyle 3x^2 + 22x + 7$
$\displaystyle 3x^2 - 7x + 2$
Oh...so because the x^2 has a coefficient of 2, in factorising I should be multiplying the second coefficient by 2...
If that makes sense. I think I understand.
(2x - 2)(x + 3)
Since -2 + 3 = 1, I figured I had it done. It matched the x coefficient. But I should've been doing -2 + 2*3 = 4, which is incorrect.
And then for (2x - 3)(x + 2), I would do 2*2 - 3 which = 1, which is the correct coefficient.
Thanks very much, I'm gunna go write that down.