# Homework Help: Simple factoring question here: Thanks!

1. Aug 30, 2012

### nukeman

1. The problem statement, all variables and given/known data

Ok, just doing a re-fresher here!

2x^2 - 7x - 2

Factor it...

First, let me ask you guys, when you see this, what is the FIRST thing you think of doing?

3. The attempt at a solution

For me... And I do not think this is correct.

I would first automatically put the following:

(2x ) (x )

?

Then, all I need to do is put in the values that add to -4???

2. Aug 30, 2012

### Kolmin

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

$$x(2x-7)-2$$

3. Aug 30, 2012

### nukeman

Ok thanks for the input.

The answer from my text says:

(2x + 1) (x - 4)

4. Aug 30, 2012

### Staff: Mentor

So far, so good.
Next, you want to find two factors of -2 so that the x terms add up to -7.

Here are the possibilities:
(2x - 2) (x + 1) -- x term is -2x + 2x = 0x
(2x + 2) (x - 1) -- x term is 2x - 2x = 0x
(2x + 1) (x - 2) -- x term is x - 4x = -3x
(2x - 1) (x + 2) -- x term is -x + 4x = 3x

So, no good, since none of the combinations gave us an x term with a coefficient of -7.

Your quadratic trinomial doesn't factor into "nice" factors (i.e., factors with integer coefficients).

5. Aug 30, 2012

### Staff: Mentor

???
This is not what it means to factor a trinomial.

6. Aug 30, 2012

### Staff: Mentor

There's a mistake somewhere. If you expand this product, you get 2x2 - 7x - 4. In post 1 you wrote 2x2 - 7x - 2. Did you make a mistake or is it a typo in your book?

7. Aug 30, 2012

### nukeman

Ohh CRAP!!! Sorry, dam!

Its: 2x^2 - 7x - 4

8. Aug 30, 2012

### Kolmin

Linguistic problems...

Sorry for the epic fail.

9. Aug 30, 2012

### Kolmin

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

$ax^2 + bx + c = a(x-u)(x-v)$

where $u$ and $v$ are the solutions of the equation.

In this case we indeed have

$$2(x-4)(x+ \frac{1}{2})$$

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

10. Aug 30, 2012

### nukeman

Don't quite understand the steps here :(

2x^2 - 7x - 4

So, like I said, I automatically put:

(2x ) (x )

Then, I need to find 2 factors of -4 that add up to -7 ?????

11. Aug 30, 2012

### Staff: Mentor

No, you need to find two factors of -4 so that the two cross terms (the terms in x) have a coefficient of -7.

You're probably thinking of a method where you multiple a times c (the coefficient of the x^2 term times the constant. Then you factor that number into factors that add up to b.

In your problem, a = 2, b = -7, and c = -4
a * c = -8

We look for factors of -8 that add to -7, and these would be 1 and -8.

Now, write the trinomial using these numbers.
2x2 + x - 8x - 4

Now form two groups:
(2x2 + x) + (-8x - 4)

Factor each group:

x(2x + 1) - 4(2x + 1)

Now, use the distributive property to write as (x - 4)(2x + 1).

12. Aug 30, 2012

### SammyS

Staff Emeritus
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.)

$\displaystyle 2x^2 - 7x - 4$
$\displaystyle =\frac{4x^2 - 7(2x)-8}{2}$

$\displaystyle =\frac{(2x)^2 - 7(2x)-8}{2}$​
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:

$\displaystyle 2x^2 - 7x - 4$
$\displaystyle =\frac{(2x)^2 - 7(2x)-8}{2}$

$\displaystyle =\frac{(2x-8)(2x+1)}{2}$

$\displaystyle =\frac{2(x-4)(2x+1)}{2}$​
Giving you the desired result.

13. Aug 30, 2012

### nukeman

Fantastic! Thank you Mark! :)

Sammy, pretty cool way of doing, never seen that!

14. Aug 31, 2012

### Mentallic

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).