Answer:
[tex]S=\left \{ 0,-1,1,2 \right \}[/tex]
Step-by-step explanation:
There are several ways to solve this quartic equation. But since the coefficients, they repeat a=1,b=2,c=1,d=2, but e=0, and they are multiple of each other, then it is more convenient to work with factoring as the method of solving it.
As if it was a quadratic one.
[tex]x^{4}-2x^{3}- x^{2} + 2x = 0\\x(x^{3}-2x^{2}-x+2)=0 \:Factoring \:out\\x[(x^{3}-2x^{2})+(-x+2)]=0 \:Grouping\\x[\mathbf{x^{2}}(x-2)+\mathbf{-1}(x+2)]=0 \:Rewriting\:the\:first\:factor\\x(x^{2}-1)(x-2)\:Expanding \:the \:first \:factor\\x(x-1)(x+1)(x-2)=0\\x=0,x=1,x=-1,x=2\\S=\left \{ 0,-1,1,2 \right \}[/tex]
Answer:
x = 0, 1, 2 and -1
Step-by-step explanation:
x⁴ − 2x³ − x² + 2x = 0.
Factorizing the equation,
x(x³ − 2x² - x + 2) = 0
⇒ x = 0 and x³ − 2x² - x + 2 = 0
⇒ x = 0 is a root of the equation
Solving for the polynomial x³ - 2x² - x + 2 = 0
When x = 1
(1)³ - 2(1)² - 1 + 2
1 - 2 - 1 + 2 = 0
Therefore, x = 1 is a root of the equation.
⇒ x - 1 = 0
Using long division approach to get the other roots
x² - x - 2
-----------------------
x - 1 ║ x³ - 2x² - x + 2
x³ - x²
---------------------
- x² - x + 2
- x² + x
------------------------
- 2x + 2
- 2x + 2
-------------------
0
We will solve the quotient x² - x - 2 to get the other roots of the equation.
Solving quadratic equation x² - x - 2 = 0 using factorization method
x² - 2x + x - 2 = 0
(x² - 2x) + (x - 2) = 0
x(x - 2) + 1(x - 2) = 0
(x - 2)(x + 1) = 0
x - 2 = 0 and x + 1 = 0
⇒ x = 2 and x = - 1 are also roots of the equations.
The solution of the equation x⁴ − 2x³ − x² + 2x = 0 is x = 0, x = 1, x = 2 and x = -1
