Anyone please help me answer this questions. Your help is very much appreciated.
Answers
Step-by-step explanation:
x4-2x3-x2-2x+1=0
One solution was found :
x ≓ 0.381966114
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((((x4) - 2x3) - x2) - 2x) + 1 = 0
Step 2 :
Polynomial Roots Calculator :
2.1 Find roots (zeroes) of : F(x) = x4-2x3-x2-2x+1
Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers
The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient
In this case, the Leading Coefficient is 1 and the Trailing Constant is 1.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1
Let us test ....
P Q P/Q F(P/Q) Divisor
-1 1 -1.00 5.00
1 1 1.00 -3.00
Polynomial Roots Calculator found no rational roots
Equation at the end of step 2 :
x4 - 2x3 - x2 - 2x + 1 = 0
Step 3 :
Quartic Equations :
3.1 Solve x4-2x3-x2-2x+1 = 0
In search of an interavl at which the above polynomial changes sign, from negative to positive or the other wayaround.
Method of search: Calculate polynomial values for all integer points between x=-20 and x=+20
Found change of sign between x= 0.00 and x= 1.00
Approximating a root using the Bisection Method :
We now use the Bisection Method to approximate one of the solutions. The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).
The function is F(x) = x4 - 2x3 - x2 - 2x + 1
At x= 1.00 F(x) is equal to -3.00
At x= 0.00 F(x) is equal to 1.00
Intuitively we feel, and justly so, that since F(x) is negative on one side of the interval, and positive on the other side then, somewhere inside this interval, F(x) is zero
Procedure :
(1) Find a point "Left" where F(Left) < 0
(2) Find a point 'Right' where F(Right) > 0
(3) Compute 'Middle' the middle point of the interval [Left,Right]
(4) Calculate Value = F(Middle)
(5) If Value is close enough to zero goto Step (7)
Else :
If Value < 0 then : Left <- Middle
If Value > 0 then : Right <- Middle
(6) Loop back to Step (3)
(7) Done!! The approximation found is Middle
Follow Middle movements to understand how it works :
Left Value(Left) Right Value(Right)
1.000000000 -3.000000000 0.000000000 1.000000000
1.000000000 -3.000000000 0.000000000 1.000000000
0.500000000 -0.437500000 0.000000000 1.000000000
0.500000000 -0.437500000 0.250000000 0.410156250
0.500000000 -0.437500000 0.375000000 0.023681641
0.437500000 -0.197250366 0.375000000 0.023681641
0.406250000 -0.084395409 0.375000000 0.023681641
0.390625000 -0.029764116 0.375000000 0.023681641
0.382812500 -0.002893683 0.375000000 0.023681641
0.382812500 -0.002893683 0.378906250 0.010430783
0.382812500 -0.002893683 0.380859375 0.003777762
0.382812500 -0.002893683 0.381835938 0.000444344
0.382324219 -0.001224093 0.381835938 0.000444344
0.382080078 -0.000389730 0.381835938 0.000444344
0.382080078 -0.000389730 0.381958008 0.000027343
0.382019043 -0.000181185 0.381958008 0.000027343
0.381988525 -0.000076919 0.381958008 0.000027343
0.381973267 -0.000024787 0.381958008 0.000027343
0.381973267 -0.000024787 0.381965637 0.000001278
0.381969452 -0.000011755 0.381965637 0.000001278
0.381967545 -0.000005238 0.381965637 0.000001278
Next Middle will get us close enough to zero:
F( 0.381966114 ) is -0.000000351
The desired approximation of the solution is:
x ≓ 0.381966114
Note, ≓ is the approximation symbol
One solution was found :
x ≓ 0.381966114
Processing ends successfully
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