Simple factoring question here: Thanks!

  • Thread starter nukeman
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  • #1
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Homework Statement



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?


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???
 

Answers and Replies

  • #2
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[tex]2x^2 - 7x - 2[/tex]
Factor it...

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

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]
 
  • #3
655
0
Ok thanks for the input.

The answer from my text says:

(2x + 1) (x - 4)
 
  • #4
34,687
6,393

Homework Statement



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?


The Attempt at a Solution



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

I would first automatically put the following:

(2x ) (x )

?
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).

Then, all I need to do is put in the values that add to -4???
 
  • #5
34,687
6,393
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]
???
This is not what it means to factor a trinomial.
 
  • #6
34,687
6,393
Ok thanks for the input.

The answer from my text says:

(2x + 1) (x - 4)
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
655
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Ohh CRAP!!! Sorry, dam!

Its: 2x^2 - 7x - 4
 
  • #8
66
0
???
This is not what it means to factor a trinomial.
:redface:

Linguistic problems...

Sorry for the epic fail.
 
  • #9
66
0
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.
 
  • #10
655
0
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
34,687
6,393
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
SammyS
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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 ?????
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.
 
  • #13
655
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Fantastic! Thank you Mark! :)

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

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).
 
  • #14
Mentallic
Homework Helper
3,798
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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
 

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