x^3 -x^2 +x -1 = 0
x^3 -x +3 -3x^2 = 0
10 points if you can solve this :?
2008-06-15 2:51 pm
回答 (13)
2008-06-15 3:05 pm
✔ 最佳答案
For the first one, you first factor out x-1, making your equation:(x-1)(x^2+1) = 0.
Because you can't easily factor x^2 + 1, just leave it as is.
Now solve for x-1 = 0 and x^2+1 = 0
So, x-1 =0
x = 1, and
x^2+1 = 0
x^2 = -1
x = sqrt(-1) = i
So x = 1, i
For the second one,
x^3-x+3-3x^2 = 0, you should re-write it so that your exponents decrease from left to right, so the equation is:
x^3 - 3x^2-x+3 = 0.
Factor out (x-3) and you get:
(x-3)(x^2-1)=0.
Factor again to get:
(x-3)(x+1)(x-1) = 0.
Now solve each one:
x-3 = 0
x = 3;
x+1 = 0
x = -1;
x-1 = 0
x=1;
So x = 3,-1, and 1.
Hope this helps
2008-06-15 10:12 pm
1)
x^3 - x^2 + x - 1 = 0
(x^2 + 1)(x - 1) = 0
x^2 + 1 = 0
x^2 = -1
x = 屉-1 (imaginary number)
x - 1 = 0
x = 1
â´ x = 1
= = = = = = = =
2)
x^3 - x + 3 - 3x^2 = 0
x^3 - 3x^2 - x + 3 = 0
(x^2 - 1)(x - 3) = 0
(x + 1)(x - 1)(x - 3) = 0
x + 1 = 0
x = -1
x - 1 = 0
x = 1
x - 3 = 0
x = 3
ⴠx = ±1 , 3
x^3 - x^2 + x - 1 = 0
(x^2 + 1)(x - 1) = 0
x^2 + 1 = 0
x^2 = -1
x = 屉-1 (imaginary number)
x - 1 = 0
x = 1
â´ x = 1
= = = = = = = =
2)
x^3 - x + 3 - 3x^2 = 0
x^3 - 3x^2 - x + 3 = 0
(x^2 - 1)(x - 3) = 0
(x + 1)(x - 1)(x - 3) = 0
x + 1 = 0
x = -1
x - 1 = 0
x = 1
x - 3 = 0
x = 3
ⴠx = ±1 , 3
2008-06-15 10:12 pm
x^3 -x^2 +x -1 = 0
By trial and error, 1 is a root , so (x-1) is a factor. That's the only solution.
x^3-3x^2-x+3=0
There are 3 roots. If you graph the function, you can see where they lie.
The roots are -1,1, and 3.
By trial and error, 1 is a root , so (x-1) is a factor. That's the only solution.
x^3-3x^2-x+3=0
There are 3 roots. If you graph the function, you can see where they lie.
The roots are -1,1, and 3.
2008-06-15 10:07 pm
for the first one: x^3-x^2+x-1=0
x^2(x-1)+x-1=0
(x-1)(x^2-1)=0
the solutions r then 1 and -1
for the second one:
x^3-x+3-3x^2=0
x(x^2-1)-3(-1+x^2)=0
(x^2-1)(x-3)=0
the solution r then 1; -1 and 3!
good luck!!!
x^2(x-1)+x-1=0
(x-1)(x^2-1)=0
the solutions r then 1 and -1
for the second one:
x^3-x+3-3x^2=0
x(x^2-1)-3(-1+x^2)=0
(x^2-1)(x-3)=0
the solution r then 1; -1 and 3!
good luck!!!
2008-06-15 10:06 pm
the solution is basic factoring by grouping:
1. x^3-x^2+x-1
i grouped the first 2 and last 2 terms:
(x^3-x^2) + (x-1)
then i factor x^2 from the first term:
x^2(x-1) + (x-1)
factor out x-1
(x^2+1)(x-1) = 0
equate each factor to 0
x^2+1 = 0
x^2 = -1
not possible
x-1=0
x=1
the answer is 1.
2. same for number 2
x^3 - x + 3 - 3x^2
i'll rearrange the terms
x^3- x - 3x^2 + 3
then group (be careful of the negative sign)
(x^3-x) - (3x^2 - 3) *take note, it becomes -3, because it is enclosed in a parenthesis preceded by a -
then i factor out
x(x^2-1) - 3(x^2-1)
factor again
(x-3)(x^2-1)
we can factor the second term too
(x-3)(x+1)(x-1)
equate each to 0
x-3=0
x=3
x+1=0
x=-1
x-1=0
x=1
so the answers are 3, 1, -1. =)
1. x^3-x^2+x-1
i grouped the first 2 and last 2 terms:
(x^3-x^2) + (x-1)
then i factor x^2 from the first term:
x^2(x-1) + (x-1)
factor out x-1
(x^2+1)(x-1) = 0
equate each factor to 0
x^2+1 = 0
x^2 = -1
not possible
x-1=0
x=1
the answer is 1.
2. same for number 2
x^3 - x + 3 - 3x^2
i'll rearrange the terms
x^3- x - 3x^2 + 3
then group (be careful of the negative sign)
(x^3-x) - (3x^2 - 3) *take note, it becomes -3, because it is enclosed in a parenthesis preceded by a -
then i factor out
x(x^2-1) - 3(x^2-1)
factor again
(x-3)(x^2-1)
we can factor the second term too
(x-3)(x+1)(x-1)
equate each to 0
x-3=0
x=3
x+1=0
x=-1
x-1=0
x=1
so the answers are 3, 1, -1. =)
2008-06-15 10:06 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 = 0 x - 1 = 0
x^2 = -1 x = 1
x = +/-1
x^3 - x + 3 - 3x^2 = 0
x^3 - x - 3x^2 + 3 = 0
x ( x^2 - 1 ) - 3 ( x^2 - 1 ) = 0
( x^2 - 1 ) ( x - 3 ) = 0
x^2 - 1 = 0 x - 3 = 0
x^2 = 1 x = 3
x = +/-1
x^2( x - 1) 1 ( x - 1 ) =0
( x^2 + 1 ) (x - 1 ) = 0
x^2 + 1 = 0 x - 1 = 0
x^2 = -1 x = 1
x = +/-1
x^3 - x + 3 - 3x^2 = 0
x^3 - x - 3x^2 + 3 = 0
x ( x^2 - 1 ) - 3 ( x^2 - 1 ) = 0
( x^2 - 1 ) ( x - 3 ) = 0
x^2 - 1 = 0 x - 3 = 0
x^2 = 1 x = 3
x = +/-1
2008-06-15 10:05 pm
x^3 -x^2 +x -1 = 0
factor by grouping
x^2(x-1) + 1(x-1)=0
(x-1)(x^2+1)=0
by the zero property either x-1=0===> x=+1
or (x^2+1)=0 = x^2 = -1 = i^2 so x= +-i
x^3 -x +3 -3x^2 = 0
x(x^2-1) -3(x^2 -1) =0
(x-3)(x^2 -1) = 0
( x-3)(x -1)(x+1) = 0
by the zero property x-3=0 ===> x=+3
or x-1=0 ===> x = +1
or x+1=0 ===> x = -1
factor by grouping
x^2(x-1) + 1(x-1)=0
(x-1)(x^2+1)=0
by the zero property either x-1=0===> x=+1
or (x^2+1)=0 = x^2 = -1 = i^2 so x= +-i
x^3 -x +3 -3x^2 = 0
x(x^2-1) -3(x^2 -1) =0
(x-3)(x^2 -1) = 0
( x-3)(x -1)(x+1) = 0
by the zero property x-3=0 ===> x=+3
or x-1=0 ===> x = +1
or x+1=0 ===> x = -1
參考: i'm a math teacher
2008-06-15 10:04 pm
x^3 -x^2 +x -1 = 0
Group
(x^3 -x^2) +(x -1) = 0
x^2 (x -1) +1(x -1) = 0
(x^2 + 1)(x - 1) = 0
Done factoring if you are restricted to real numbers, otherwise:
(x + i)(x - i)(x - 1)
x = 1, i, or -1
x^3 -x +3 -3x^2 = 0
Reorder and Group
x^3 -3x^2 -x +3 = 0
(x^3 - 3x^2) - (x - 3) = 0
x^2(x - 3) - 1(x - 3) = 0
(x^2 - 1)(x - 3) = 0
(x + 1)(x - 1)(x - 3) = 0
x = -1, 1, or 3
Group
(x^3 -x^2) +(x -1) = 0
x^2 (x -1) +1(x -1) = 0
(x^2 + 1)(x - 1) = 0
Done factoring if you are restricted to real numbers, otherwise:
(x + i)(x - i)(x - 1)
x = 1, i, or -1
x^3 -x +3 -3x^2 = 0
Reorder and Group
x^3 -3x^2 -x +3 = 0
(x^3 - 3x^2) - (x - 3) = 0
x^2(x - 3) - 1(x - 3) = 0
(x^2 - 1)(x - 3) = 0
(x + 1)(x - 1)(x - 3) = 0
x = -1, 1, or 3
2008-06-15 10:01 pm
x^3-x^2+x-1=0
=>x^2(x-1)+1(x-1)=0
=>(x-1)(x^2+1)=0
hence, (x-1)=0 or (x^2+1)=0
(x-1)=0 =>x=1
(x^2+1)=0 =>x^2=-1[not possible coz square of a number can't be negetive]
the solution is: x=1
x^3-x+3-3x^2=0
=>x^3-3x^2-x+3=0
=>x^2(x-3)-1(x-3)=0
hence, (x^2+1)=0 or (x-3)=0
(x^2+1)=0 =>x^2 =-1[not possible (same reason as i mentioned above)]
(x-3)=0
=>x=3
so the solution: x=3
=>x^2(x-1)+1(x-1)=0
=>(x-1)(x^2+1)=0
hence, (x-1)=0 or (x^2+1)=0
(x-1)=0 =>x=1
(x^2+1)=0 =>x^2=-1[not possible coz square of a number can't be negetive]
the solution is: x=1
x^3-x+3-3x^2=0
=>x^3-3x^2-x+3=0
=>x^2(x-3)-1(x-3)=0
hence, (x^2+1)=0 or (x-3)=0
(x^2+1)=0 =>x^2 =-1[not possible (same reason as i mentioned above)]
(x-3)=0
=>x=3
so the solution: x=3
2008-06-15 10:05 pm
you can do it by horner method, but im sorry i cant explain it by typing!
1. First, find the factors of -1.
They are -1 and 1. Then divide the function by -1 or 1. If the remainder is 0, then it is the root of the function.
it is found that the ROOT of the first function is 1. The quotient is x^2 +1, as it can't be factorized, it is the FACTOR of the function.
2. Find the factors of 3 which are -1, 1, -3, and 3.
I found the first root is -1 and the quotient is x^2 - 4x +3
Factorize it and you will find the other roots which are 1 and 3.
1. First, find the factors of -1.
They are -1 and 1. Then divide the function by -1 or 1. If the remainder is 0, then it is the root of the function.
it is found that the ROOT of the first function is 1. The quotient is x^2 +1, as it can't be factorized, it is the FACTOR of the function.
2. Find the factors of 3 which are -1, 1, -3, and 3.
I found the first root is -1 and the quotient is x^2 - 4x +3
Factorize it and you will find the other roots which are 1 and 3.
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