how to factor x^3+x^2-x-1?
回答 (12)
2016-02-17 1:32 am
How to factor x^3+x^2-x-1?
Factor by grouping...
(x³ + x²) - (x + 1) = x²(x + 1) - (x + 1) = (x + 1)(x² - 1)
factor the DIFFERENCE OF SQUARES...
(x + 1)(x + 1)(x - 1) = (x + 1)² (x - 1)
check by multiplying...do something !
EDIT: thumbs down for the correct answer...UP YOURS !
Factor by grouping...
(x³ + x²) - (x + 1) = x²(x + 1) - (x + 1) = (x + 1)(x² - 1)
factor the DIFFERENCE OF SQUARES...
(x + 1)(x + 1)(x - 1) = (x + 1)² (x - 1)
check by multiplying...do something !
EDIT: thumbs down for the correct answer...UP YOURS !
2016-02-17 1:40 am
for this problem, you can factor by grouping:
(x^3+x^2) (-x-1)
Take out x^2 from first parenthesis leaving you with x^2(x+1)
Take out a -1 from the second parenthesis leaving you with -1(x+1)
Combine to result in (x^2-1) (x+1)
Factor the first parenthesis using difference of squares resulting in (x-1)(x+1)
The final answer is (x-1)(x+1)(x+1) or (x+1)^2 (x-1)
(x^3+x^2) (-x-1)
Take out x^2 from first parenthesis leaving you with x^2(x+1)
Take out a -1 from the second parenthesis leaving you with -1(x+1)
Combine to result in (x^2-1) (x+1)
Factor the first parenthesis using difference of squares resulting in (x-1)(x+1)
The final answer is (x-1)(x+1)(x+1) or (x+1)^2 (x-1)
2016-02-20 7:52 pm
You split the two terms.
x^3 + x^2. Now take the greatest common factor, x^2. Now you have x^2 (x+1)
Then, take -1 out of -x-1. Then you have the factors (x^2-1) (x+1).
Your roots are 1, 1, and -1.
x^3 + x^2. Now take the greatest common factor, x^2. Now you have x^2 (x+1)
Then, take -1 out of -x-1. Then you have the factors (x^2-1) (x+1).
Your roots are 1, 1, and -1.
2016-02-19 5:08 am
x²(x+1)-(x+1) = (x+1)(x²-1) = (x+1)(x+1)(x-1) = (x+1)²(x-1) <== ans.
2016-02-18 7:44 pm
The solution of this question x^3 + x^2 - x - 1
= x^2(x + 1) - (x + 1)
= (x^2 - 1)(x + 1)
= (x + 1)^2(x - 1)
= x^2(x + 1) - (x + 1)
= (x^2 - 1)(x + 1)
= (x + 1)^2(x - 1)
2016-02-18 1:57 am
ans.=(x+1)^2,(x-1)
checking:
=(x+1)(x+1)
=x^2+2x+1
=(x^2+2x+1)(x-1)
=x^3+x^2-x-1
checking:
=(x+1)(x+1)
=x^2+2x+1
=(x^2+2x+1)(x-1)
=x^3+x^2-x-1
2016-02-17 2:48 pm
x^3+x^2-x-1=0
x^2(x+1)-1(x+1)=0
(x^2-1)(x+1)=0
x^2=1,x=-1
x=1,x=-1(twice)
x^2(x+1)-1(x+1)=0
(x^2-1)(x+1)=0
x^2=1,x=-1
x=1,x=-1(twice)
2016-02-17 8:58 am
The easiest way to factor any polynomial is to look for patterns and group terms. In this case I can see the expression x^3+x^2-x-1 is easy to factor.
x^3+x^2-x-1 = x^2(x+1)-(x+1) = (x^2-1)(x+1)
x^2 - 1 is the difference of two squares which means (x^2-1) = (x+1)(x-1)
We know this because x^2-1 = x^2 + x- x- 1 = x(x+1) - (x+1) = (x+1)(x-1) (x - x = 0)
So x^3+x^2-x-1 = (x^2-1)(x+1) = (x+1)(x-1)(x+1) = (x-1)(x+1)^2
x^3+x^2-x-1 = x^2(x+1)-(x+1) = (x^2-1)(x+1)
x^2 - 1 is the difference of two squares which means (x^2-1) = (x+1)(x-1)
We know this because x^2-1 = x^2 + x- x- 1 = x(x+1) - (x+1) = (x+1)(x-1) (x - x = 0)
So x^3+x^2-x-1 = (x^2-1)(x+1) = (x+1)(x-1)(x+1) = (x-1)(x+1)^2
2016-02-17 8:21 am
x^3 + x^2-x-1
=x^2 ( x + 1) -1( x+ 1)
= ( x+ 1 )( x^2 - 1)
= ( x+ 1 ) ( x+ 1 ) ( x- 1 )
= ( x+ 1 )^2 ( x- 1 ) ANSWER
=x^2 ( x + 1) -1( x+ 1)
= ( x+ 1 )( x^2 - 1)
= ( x+ 1 ) ( x+ 1 ) ( x- 1 )
= ( x+ 1 )^2 ( x- 1 ) ANSWER
2016-02-17 2:54 am
x^3 + x^2 - x - 1
= x^2(x + 1) - (x + 1)
= (x^2 - 1)(x + 1)
= (x + 1)^2(x - 1)
= x^2(x + 1) - (x + 1)
= (x^2 - 1)(x + 1)
= (x + 1)^2(x - 1)
2016-02-17 1:33 am
You could factor an x2 out of the first two terms and a -1 out of the second two terms:
x2(x + 1)-1(x + 1)
You can then factor out the (x + 1) out of both terms:
(x2 -1)(x + 1)
Then notice that (x2 - 1) is a difference of squares factorable into (x -1)(x + 1)
So your final answer should be (x - 1)(x + 1)( x + 1)
x2(x + 1)-1(x + 1)
You can then factor out the (x + 1) out of both terms:
(x2 -1)(x + 1)
Then notice that (x2 - 1) is a difference of squares factorable into (x -1)(x + 1)
So your final answer should be (x - 1)(x + 1)( x + 1)
2016-02-17 1:31 am
Hint: Divide by (x-1)
收錄日期: 2021-04-21 16:53:43
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