I am having the most difficult time factoring this equation out. I should have paid more attention in precalculus in high school :(
$2x^3 + 3x^2 - 2x = x(2x^2 + 3x - 2) = x(2x-1)(x+2)$
How do you get the $(2x-1)(x+2)$ out of $(2x^2+3x-2)$?
Quadratic equation didn't work, I can't take out a common variable, I am out of options in my head
On
You can always use the quadratic equation to factor a quadratic polynomial $Q(x)$.
If the solutions to $Q(x)=0$ are $x=r$ and $x=s$, then the polynomial can be factored as $$Q(x)=k(x-r)(x-s)$$ where $k$ is the leading coefficient of $Q(x)$.
In your case, $Q(x)=2x^2+3x-2$ and the quadratic formula says the solutions to $Q(x)=0$ are $$x=\frac{-3 \pm\sqrt{3^2-4(2)(-2)}}{2(2)}=\frac{-3 \pm 5}{4}$$ So the solutions are $x=\frac12$ and $x=-2$.
This means $Q(x)$ factors as $2(x-\frac12)(x+2)$. It is convenient to combine the leading constant with the first factor to get rid of the fraction, giving $$\boxed{Q(x)=(2x-1)(x+2)}.$$
It may be worth remembering how the quadratic formula is derived in the first place. Believe it or not, it's through factoring (!). Here's how. Suppose we have a quadratic equation (assume $a\neq0$, otherwise it's really just a linear equation):
$$ax^2+bx+c=0$$
We work towards the goal of building the constant discriminant $b^2-4ac$ on the right side:
$$ax^2+bx = \boxed{-c}\tag{subtract $c$ from both sides}$$ $$4a^2x^2 + 4abx = \boxed{-4ac}\tag{multiply both sides by $4a$}$$ $$4a^2x^2 + 4abx +b^2= \boxed{b^2-4ac}\tag{add $b^2$ to both sides}$$
When you do this, the left side is always a perfect square polynomial (the square of a binomial). You can then factor the left side: $$\overbrace{(2ax+b)(2ax+b)}^{4a^2x^2 + 4abx +b^2\textrm{ factored}}= \overbrace{b^2-4ac}^{\textrm{constant discriminant}}$$ Assuming that $b^2-4ac$ isn't negative, we can write $b^2-4ac=(\sqrt{b^2-4ac})^2$ so $$(2ax+b)^2 = (\sqrt{b^2-4ac})^2$$ $$(2ax+b)^2 - (\sqrt{b^2-4ac})^2 = 0$$ which can be factored, being a difference of squares, as $$(2ax+b - \sqrt{b^2-4ac})(2ax+b + \sqrt{b^2-4ac})=0$$ Dividing both sides by $2a$ twice, and then multiplying by $a$, we have $$a\left(x+\frac{b- \sqrt{b^2-4ac}}{2a}\right)\left(x+\frac{b+ \sqrt{b^2-4ac}}{2a}\right)=0$$ $$a\left(x-\left[\frac{-b+ \sqrt{b^2-4ac}}{2a}\right]\right)\left(x-\left[\frac{-b- \sqrt{b^2-4ac}}{2a}\right]\right)=0$$ The left side is indeed of the form $k(x-r)(x-s)$, where $k$ is the leading coefficient of $Q(x)$ (that is, $a$) and $r,s$ are the solutions $\tfrac{-b\pm \sqrt{b^2-4ac}}{2a}$from the quadratic formula.
If the discriminant is negative, the same formula works but the square root will produce an imaginary number. The quadratic will still factor, but the factors will contain complex numbers. In other words, if $b^2-4ac<0$, we can write $$\pm\sqrt{b^2-4ac} = \pm i\sqrt{-(b^2-4ac)}.$$
On
My thought process as I go to factor this expression goes something like the following:
$$2x^3+3x^2-2x$$
"Oh, there is no constant term and everything is a multiple of $x$... so that means I can safely factor that out by itself and I can look more closely at what is left"
$$x(2x^2+3x-2)$$
"Allright, I've got a quadratic left over... oh hey, I was taught about quadratics a really long time in great detail. Hmm, which technique do I want to use today. Well, I get confused if the coefficient of the $x^2$ is not one, so I'll just go and use the quadratic formula"
The Quadratic Formula: (most people are taught to memorize this, but it should be well within your ability to prove) Given a degree two polynomial of the form $Ax^2+Bx+C$, it can be factored as
$$A\left(x-\frac{-B+\sqrt{B^2-4AC}}{2A}\right)\left(x+\frac{-B-\sqrt{B^2-4AC}}{2A}\right)$$
So, looking at $2x^2+3x-2$ in this light we see that this section factors as
$$2\left(x-\dfrac{-3+\sqrt{3^2-4\cdot 2\cdot (-2)}}{2\cdot 2}\right)\left(x-\dfrac{-3-\sqrt{3^2-4\cdot 2\cdot (-2)}}{2\cdot 2}\right)$$
which after simplifying all of the arithmetic and including the $x$ we left to the side earlire becomes
$$2x(x-\frac{1}{2})(x+2)$$
(or if you prefer moving the $2$ on the far outside into one of the parentheses and avoiding fractions can be written as $x(2x-1)(x+2)$)
Other techniques exist, especially when you assume the factorization includes only integers as are common in introductory examples, however the power of the quadratic formula is immense and it will always work even in those situations where other techniques might fail.
When the quadratic formula is first taught, it is common in the United States at least for it to be taught via song to make it easier to remember, for example here. (This is not the melody I was taught it to, but so long as you can fit it to a melody it shouldn't really matter which melody it is).
It is worth pointing out that there does exist a generalized formula for how to factor an arbitrary cubic equation that some people might have been taught and can memorize but it is a great deal more complicated than the quadratic formula. It is not expected for you to learn this. It is also worth pointing out that there also does exist a generalized formula for how to factor an arbitrary quartic, but this formula takes several pages to even write down. No one in their right minds would bother memorizing the fully generalized formula for the quartic, only maybe some few special cases. Finally, in your senior year of a mathematics degree in undergraduate or in graduate school it is common to reproduce the proof that there does not and cannot exist a fully generalized formula for finding the factorization of a quintic or above.
Being a high school student I can help you out. $$2x^2+3x-2$$ $$=2x^2+4x-x-2$$ $$=2x(x+2)-(x+2)$$ $$=(2x-1)(x+2)$$