1. ## Factoring Questions

1. When factoring by grouping I am having problems understanding a particular step.
Expression:$\displaystyle 2x^4-6x-x^3y+3y$

$\displaystyle =(2x^4-6x)-(x^3y+3y)$
$\displaystyle =2x(x^3-3)+y(x3-3)$ Why are we able to change the signs towards the end of the expression? In the first step its a "-" sign and at the next step its a "+" sign.
$\displaystyle =(2x-y)(x^3-3)$ and in the final solution the sign changes again...

2. Reducing a fraction with -1
"Reducing a fraction or adding two fractions sometimes only requires that -1 be factored from one or more denominators"
Expression:$\displaystyle \frac{3}{y-x}+\frac{x}{x-y}=\frac{-3+x}{x-y}$
-The book lists the steps to solve this but they are rather confusing and they do not clearly explain, I would prefer a member on here run me through the above expression and provide the how/why when solving this problem.

2. 1) If you divide a positive (+3y) by a negative (-y), what should be the sign on the answer?

2) You have y - x, which equals -x + y, which equals -1(x - y), which then allows you to have a common denominator.

3. Originally Posted by stapel
1) If you divide a positive (+3y) by a negative (-y), what should be the sign on the answer?

2) You have y - x, which equals -x + y, which equals -1(x - y), which then allows you to have a common denominator.

im afraid I do not understand your first explanation, the second is more clear to me at the moment.

4. Originally Posted by allyourbass2212
1. When factoring by grouping I am having problems understanding a particular step.
Expression:$\displaystyle 2x^4-6x-x^3y+3y$
$\displaystyle =(2x^4-6x)-(x^3y+3y)$
This is wrong. It should be $\displaystyle =(2x^4-6x)-(x^3y\;{\color{red}-}\;3y)$ The reason is that we essentially factored out a -1 in the last two terms.

$\displaystyle =2x(x^3-3)+y(x3-3)$
This should be $\displaystyle =2x(x^3-3)\;{\color{red}-}\;y(x^3 - 3)$

$\displaystyle =(2x-y)(x^3-3)$
This is correct.

01

5. Originally Posted by yeongil
This is wrong. It should be $\displaystyle =(2x^4-6x)-(x^3y\;{\color{red}-}\;3y)$ The reason is that we essentially factored out a -1 in the last two terms.
This should be $\displaystyle =2x(x^3-3)\;{\color{red}-}\;y(x^3 - 3)$

Would you mine explaining this in further detail? Because I still dont understand where/why/ and how the -1 is factored out for the last 2 terms .

6. Originally Posted by allyourbass2212
This is wrong. It should be $\displaystyle =(2x^4-6x)-(x^3y\;{\color{red}-}\;3y)$ The reason is that we essentially factored out a -1 in the last two terms.
Would you mine explaining this in further detail? Because I still dont understand where/why/ and how the -1 is factored out for the last 2 terms .
It's because of the parentheses.

If you take an expression like
-a - b
you can factor out a -1. In doing so, you switch the signs inside:
-1(a + b)
but you don't need to write the 1 really:
-(a + b)

Now if you have an expression like
-ax + bx
Factor out a -1x (or -x) and switch the signs inside:
-1x(a - b) = -x(a - b)

The original expression was
$\displaystyle 2x^4-6x-x^3y+3y$

We're factoring by grouping. Factoring out the common factor from the first two terms is easy enough:
$\displaystyle 2x(x^3-3)-x^3y+3y$

We factor out a -1y (or a -y) from the last two terms:
$\displaystyle 2x(x^3-3)\;{\color{red}-1y}\;(x^3-3)$
$\displaystyle 2x(x^3-3)\;{\color{red}-y}\;(x^3-3)$

01

7. Many Thanks Kind Sir!

Yeongil, may I also ask that you clarify my expression 2 question?

2. Reducing a fraction with -1
Expression:$\displaystyle \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-y+x)}+\frac{x}{x-y}= \frac{3}{-(x-y)}+\frac{x}{x-y}$, now the last step here is where I become completely lost, where the denominator goes from $\displaystyle \frac{3}{-(-y+x)}$ to $\displaystyle \frac{3}{-x(x-y)}$ in the next step in solving the expression

then the last part I am also lost
$\displaystyle =\frac{3}{-(x-y)}+\frac{x}{x-y}=\frac{-3}{x-y}+\frac{x}{x-y}$ in the last part of solving this expression how does the once negative denominator of $\displaystyle =\frac{3}{-(x-y)}$ go to $\displaystyle \frac{-3}{x-y}$ in the next part of the solution?

8. $\displaystyle \frac{3}{y - x} + \frac{x}{x - y} = \frac{-3 + x}{x - y}$

I just find it easy to remember this property:
a - b = -(b - a)

So the denominator of the first fraction would be:
y - x = -(x - y)

Anyway...
$\displaystyle \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-({\color{red}-y+x})}+\frac{x}{x-y}= \frac{3}{-({\color{red}x-y})}+\frac{x}{x-y}$
You're talking about the changes in red? That's the commutative property of addition: a + b = b + a. So inside the parentheses:
-(-y + x) = -(x + (-y)) = -(x - y). You switch the x and -y.

$\displaystyle =\frac{3}{-(x-y)}+\frac{x}{x-y}=\frac{-3}{x-y}+\frac{x}{x-y}$ in the last part of solving this expression how does the once negative denominator of $\displaystyle =\frac{3}{-(x-y)}$ go to $\displaystyle \frac{-3}{x-y}$ in the next part of the solution?
You should also know that if you have a fraction that is negative, then the following are equivalent:
$\displaystyle -\frac{a}{b} = \frac{-a}{b} = \frac{a}{-b}$

We have to use this property because here:
$\displaystyle \frac{3}{{\color{red}-}(x-y)}+\frac{x}{x-y}$
the denominators are still not the same -- the first fraction has that negative sign outside the parenthesis. No problem -- the fraction property I stated above tells me that I can just move the negative to the numerator and the fraction is equivalent.
$\displaystyle \frac{{\color{red}-}3}{(x-y)}+\frac{x}{x-y}$

Hope that helps!

01

9. Yes very much thank you so very much!