AlgebraII factoring polynomials equation: $2x^2 - 11x - 6$
Using the quadratic formula, I have found the zeros: $x_1 = 6, x_2 = -\frac{1}{2}$
Plug the zeros in: $2x^2 + \frac{1}{2}x - 6x - 6$
This is where I get lost. I factor $-6x - 6$ to: $-6(x + 1)$, but the answer says otherwise. I am also having trouble factoring the left side.
Could someone please explain to me why the answer to the question was: $(x - 6)(2x + 1)$. How does $-\frac{1}{2}$ become $1$?
 A: Multiply the first and last coefficient: 
$$2\cdot (-6) = -12$$
You want factors of $-12$ which sum to $-11$. They are $-12$ and $+1$.
$$2x^2 - {\bf 11x}-6$$
$$2x^2  {\bf -12x + 1x}-6$$
$$2x(x-6) + 1(x-6)$$
$$(2x+1)(x-6)$$

Addendum: Knowing just the roots is not enough. You have foudn the roots to be $x=-\frac{1}{2}$ and $x=6$, so the polynomial will be "similar" to
$$\left(x+\frac{1}{2}\right)(x-6)$$
Multiplying by $2$ will give you
$$(2x+1)(x-6)$$
which is the correct factorization. But it is possible to have started with $$4x^2-22x-12$$ which will have the same roots, and so this process would not produce the correct factorization of $$4x^2-22x-12=2(2x+1)(x-6)$$. What you could do, is look at the leading coefficient of the starting polynomial and realize you'll need to multiply $
\left(x+\frac{1}{2}\right)(x-6)$ by $4$ instead. This is kind of annoying, so I would stick to the above method.
A: $$
2x^2-11x - 6 = 2(x -\bullet)(x-\bullet)
$$
The two "$\bullet$"s are the two roots, $6$ and $\dfrac{-1}2$.  So you have
$$
2(x-6)\left(x-\frac {-1} 2 \right) = (x-6)\  \underbrace{2\left(x+\frac 1 2 \right)} = (x-6)(2x+1).
$$
So $2\left(x + \dfrac 1 2\right)$ becomes $2x+1$.
A: When you factor out the equation $2x^2-11x-6$, you get $(x-6)(2x+1)$ (David Peterson did the factoring process). This shows that the functions has two zeros in the graph. Thus, we have to sent $(x-6)(2x+1)$ equal to $0$: $$(x-6)(2x+1)=0.$$ Then we find the zeros: 
$$(x-6)(2x+1)=0$$ 

$$x-6=0$$ 
$$\boxed{x=6}$$ 

$$2x+1=0$$ 
$$2x=-1$$ 
$$\boxed{x=-\frac{1}{2}}.$$ 
Therefore, we can clearly see that by factoring and sovling for $x$, we get  $$\boxed{x=6,-\frac{1}{2}}.$$
A: Knowing that the solutions are $x=r$ and $x=s$, you can work backwards to obtain the original quadratic (up to a constant multiplier):
$$x=r \textrm{ or } x=s$$
$$\iff x-r=0 \textrm{ or } x-s=0$$
$$\iff (x-r)(x-s)=0$$
$$\iff k(x-r)(x-s)=0 \textrm{ for any given nonzero constant } k$$
Now all you need to do is pick the value of $k$ that scales the terms properly to match the original given quadratic. But that's easy--if the original is
$$ax^2+bx+c=0$$ then you have to pick $$k=a$$ to get the coefficient of $x^2$ to be the same in the two equations.
You could also see what it takes to get the linear term or the constant term to match, but you will get the same answer. In other words, if you require that
$$k(x-r)(x-s)$$ and $$ax^2+bx+c$$ are the same expression for all $x$ (where $a,b,c, r,s$ are known and $k$ is to be determined), then equating coefficients on both sides gives you $$k=a$$ $$-k(r+s)=b$$ $$krs=c$$ which  gives you $k$ in three apparently different ways, but they will all be the same number:
$$k=a$$ $$k=-\frac{b}{r+s}$$ $$k=\frac{c}{rs}$$ (watch out if you have division by zero in the two final versions, you can't do that).
Note that if $k$ will clear a fraction in $r$ or $s$, it is often distributed over one of the linear factors instead of leaving it out front (this is the case with your problem). Similarly, you could distribute two numbers whose product is $k$ over each of the linear factors separately.
