Question 651820: how do you factor these expressions? x to the third power + x to the second power-x-1, 8x to the third power + 4x to the second power - 2x - 1, and 2x to the third power - 3x to the second power - 2x + 3.
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website! x to the third power + x to the second power-x-1, 8x to the third power + 4x to the second power - 2x - 1
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, (x+1)
x²(x+1)-1(x+1)
which we can factor out leaving the x² and the -1 to put
in parentheses:
(x+1)(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:
(x+1)(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, (2x-3)
x²(2x-3)-1(2x-3)
which we can factor out leaving the x² and the -1 to put
in parentheses:
(2x-3)(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:
(2x-3)(x-1)(x+1)
Edwin
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