# Simple factoring question here: Thanks

• nukeman
In summary, Sammy provides a neat trick for solving quadratic equations by multiplying the leading term by 2, then adding the two terms.
nukeman

## 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?

nukeman said:
$$2x^2 - 7x - 2$$
Factor it...

First, let me ask you guys, when you see this, what is the FIRST thing you think of doing?
I take a look at the variable and I check where it is around.
So, here I go for:

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

Ok thanks for the input.

The answer from my text says:

(2x + 1) (x - 4)

nukeman said:

## 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).

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

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

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

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

nukeman said:
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?

Ohh CRAP! Sorry, dam!

Its: 2x^2 - 7x - 4

Mark44 said:
?
This is not what it means to factor a trinomial.

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:

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

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.

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 ?

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).

nukeman said:
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.)

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

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

Fantastic! Thank you Mark! :)

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

Mark44 said:
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).

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).

## 1. What is factoring?

Factoring is a process in mathematics where a polynomial expression is broken down into smaller, simpler expressions. This can make solving equations and finding roots easier.

## 2. How do I know when to use factoring?

Factoring is often used when solving equations or finding the roots of a polynomial expression. If you notice a common factor in each term of the expression, factoring may be a useful strategy.

## 3. What is the difference between factoring and simplifying?

Factoring and simplifying are similar processes, but they are not the same. Factoring is breaking down an expression into smaller expressions, while simplifying is reducing an expression to its simplest form.

## 4. Can factoring be used for all types of expressions?

No, factoring can only be used for certain types of expressions, specifically polynomial expressions. Other types of expressions, such as trigonometric or exponential expressions, require different methods of simplification.

## 5. How can factoring be useful in real life?

Factoring can be useful in real life for solving problems in areas such as business, finance, and engineering. It can also be helpful in understanding patterns and relationships in data analysis and statistics.

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