Simple factoring question here: Thanks!

I take a look at the variable and I check where it is around.
So, here I go for:

[tex]x(2x-7)-2[/tex]

 

Linguistic problems...

Sorry for the epic fail.

 

Btw, having now an idea about the problem, I go for the solution of the corresponding equation and then I follow this principle:

[itex]ax^2 + bx + c = a(x-u)(x-v)[/itex]

where [itex]u[/itex] and [itex]v[/itex] are the solutions of the equation.

In this case we indeed have

[tex]2(x-4)(x+ \frac{1}{2})[/tex]

which is equal to the result posted by OP with some manipulation.

 

nukeman,

Here's a little trick that I actually picked up from some of my students:

Multiply by 2 to make the leading term a perfect square. (In the en you must divide the result by 2, but that will work fine, because both terms in one of the factors will be multiples of 2.)

[itex]\displaystyle 2x^2 - 7x - 4[/itex]

[itex]
\displaystyle =\frac{4x^2 - 7(2x)-8}{2}[/itex]

[itex]\displaystyle =\frac{(2x)^2 - 7(2x)-8}{2}[/itex]​

Now threat the 2x like it is some other variable such as u.

u² - 7u - 8 = (u -  ?  )(u +  ?  )

Of course those should be filled in with 8 and 1 .

u² - 7u - 8 = (u - 8)(u + 1)

Then back to the original:

[itex]\displaystyle 2x^2 - 7x - 4[/itex]

[itex]
\displaystyle =\frac{(2x)^2 - 7(2x)-8}{2}[/itex]

[itex]\displaystyle =\frac{(2x-8)(2x+1)}{2}[/itex]

[itex]\displaystyle =\frac{2(x-4)(2x+1)}{2}[/itex]​

Giving you the desired result.

 

Learn the cross method, it'll save you a heap of heartache. I still remember it even after years of not using it (don't be discouraged by that, it's just because most the quadratics I see can't be factorized with integer coefficients).

http://www.youtube.com/watch?v=2dqAPKgC-KI