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This is really urgent.... if you can answer this quickly please do so, would appreciate it so much.

Using the process

1. Look for GCF

2. Look for a difference of two squares

3. Use trial and error

4. Group

Please walk me through each step using this process and an easy way for me to answer the next 40 problems I have on this

Problem 1: y^{4} - 2y^{2} - y^{3}

Problem 2: x^{2} - xy - x + y

MrPatel Nov 6, 2017

#1**+3 **

**1.** y^{4} - 2y^{2} - y^{3}

First we want to look for something that is common to all the terms...we want to take out the GCF . In this case, we can take out y^{2} from each term, like this...

= y^{2}( y^{2} - 2 - y ) Notice how distributing y^{2} gets us back to the original expression.

= y^{2}( y^{2} - y - 2 )

There are three terms remaining...we cannot factor it as a difference of squares.

( A difference of squares is always in the form a^{2} - b^{2} , just two terms. )

So we need to think of two numbers that add to -1 and multiply to -2 .

How about -2 and +1 ? So our expression factors like this...

= y^{2}( y - 2 )( y + 1 )

**2.** x^{2} - xy - x + y Nothing is common to all the terms. Let's rearrange it like this...

= x^{2} - x - xy + y Now we can take out x from the first two terms.

= x(x - 1) - xy + y Notice that distributing the x makes the previous expression.

Now let's take out -y from the last two terms.

= x(x - 1) - y(x - 1) Again, check this with distrubuting.

Now...there is a common factor in the remaining terms. It is (x - 1) . Let's factor that out.

= (x - 1)(x - y)

hectictar Nov 6, 2017

#1**+3 **

Best Answer

**1.** y^{4} - 2y^{2} - y^{3}

First we want to look for something that is common to all the terms...we want to take out the GCF . In this case, we can take out y^{2} from each term, like this...

= y^{2}( y^{2} - 2 - y ) Notice how distributing y^{2} gets us back to the original expression.

= y^{2}( y^{2} - y - 2 )

There are three terms remaining...we cannot factor it as a difference of squares.

( A difference of squares is always in the form a^{2} - b^{2} , just two terms. )

So we need to think of two numbers that add to -1 and multiply to -2 .

How about -2 and +1 ? So our expression factors like this...

= y^{2}( y - 2 )( y + 1 )

**2.** x^{2} - xy - x + y Nothing is common to all the terms. Let's rearrange it like this...

= x^{2} - x - xy + y Now we can take out x from the first two terms.

= x(x - 1) - xy + y Notice that distributing the x makes the previous expression.

Now let's take out -y from the last two terms.

= x(x - 1) - y(x - 1) Again, check this with distrubuting.

Now...there is a common factor in the remaining terms. It is (x - 1) . Let's factor that out.

= (x - 1)(x - y)

hectictar Nov 6, 2017