Question 651820
x to the third power + x to the second power-x-1, 8x to the third power + 4x to the second power - 2x - 1
<pre>
x³ + x² - x - 1

Factor the first two terms x³ + x² by taking out the
greatest common factor, x². getting x²(x+1)

Factor the last two terms -x-1 by taking out -1, 
getting -1(x+1)

So we have

x²(x+1)-1(x+1)

Notice that there is a common factor, <font color="red">(x+1)</font>

x²<font color="red">(x+1)</font>-1<font color="red">(x+1)</font>

which we can factor out leaving the x² and the -1 to put 
in parentheses:

<font color="red">(x+1)</font>(x²-1)

Now we notice that the (x²-1) can be factored as the difference
of two squares as (x-1)(x+1), so we now have:

<font color="red">(x+1)</font>(x-1)(x+1)

Since two of the factors are the same (x+1), we write the final answer
with that factor squared:

(x+1)²(x-1)

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2x to the third power - 3x to the second power - 2x + 3. 

2x³ - 3x² - 2x + 3

Factor the first two terms 2x³ - 3x² by taking out the
greatest common factor, x². getting x²(2x-3)

Factor the last two terms -2x+3 by taking out -1, 
getting -1(2x-3)

So we have

x²(2x-3)-1(2x-3)

Notice that there is a common factor, <font color="red">(2x-3)</font>

x²<font color="red">(2x-3)</font>-1<font color="red">(2x-3)</font>

which we can factor out leaving the x² and the -1 to put 
in parentheses:

<font color="red">(2x-3)</font>(x²-1)

Now we notice that the (x²-1) can be factored as the difference
of two squares as (x-1)(x+1), so we end up with:

<font color="red">(2x-3)</font>(x-1)(x+1)

Edwin</pre>