Solution - Finding the roots of polynomials
Other Ways to Solve
Finding the roots of polynomialsStep by Step Solution
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
x^10-(1024)=0
Step by step solution :
Step 1 :
Trying to factor as a Difference of Squares :
1.1 Factoring: x10-1024
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 1024 is the square of 32
Check : x10 is the square of x5
Factorization is : (x5 + 32) • (x5 - 32)
Polynomial Roots Calculator :
1.2 Find roots (zeroes) of : F(x) = x5 + 32
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 32.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,8 ,16 ,32
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 31.00 | ||||||
| -2 | 1 | -2.00 | 0.00 | x + 2 | |||||
| -4 | 1 | -4.00 | -992.00 | ||||||
| -8 | 1 | -8.00 | -32736.00 | ||||||
| -16 | 1 | -16.00 | -1048544.00 | ||||||
| -32 | 1 | -32.00 | -33554400.00 | ||||||
| 1 | 1 | 1.00 | 33.00 | ||||||
| 2 | 1 | 2.00 | 64.00 | ||||||
| 4 | 1 | 4.00 | 1056.00 | ||||||
| 8 | 1 | 8.00 | 32800.00 | ||||||
| 16 | 1 | 16.00 | 1048608.00 | ||||||
| 32 | 1 | 32.00 | 33554464.00 |
The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms
In our case this means that
x5 + 32
can be divided with x + 2
Polynomial Long Division :
1.3 Polynomial Long Division
Dividing : x5 + 32
("Dividend")
By : x + 2 ("Divisor")
| dividend | x5 | + | 32 | ||||||||||
| - divisor | * x4 | x5 | + | 2x4 | |||||||||
| remainder | - | 2x4 | + | 32 | |||||||||
| - divisor | * -2x3 | - | 2x4 | - | 4x3 | ||||||||
| remainder | 4x3 | + | 32 | ||||||||||
| - divisor | * 4x2 | 4x3 | + | 8x2 | |||||||||
| remainder | - | 8x2 | + | 32 | |||||||||
| - divisor | * -8x1 | - | 8x2 | - | 16x | ||||||||
| remainder | 16x | + | 32 | ||||||||||
| - divisor | * 16x0 | 16x | + | 32 | |||||||||
| remainder | 0 |
Quotient : x4-2x3+4x2-8x+16 Remainder: 0
Polynomial Roots Calculator :
1.4 Find roots (zeroes) of : F(x) = x4-2x3+4x2-8x+16
See theory in step 1.2
In this case, the Leading Coefficient is 1 and the Trailing Constant is 16.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,8 ,16
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 31.00 | ||||||
| -2 | 1 | -2.00 | 80.00 | ||||||
| -4 | 1 | -4.00 | 496.00 | ||||||
| -8 | 1 | -8.00 | 5456.00 | ||||||
| -16 | 1 | -16.00 | 74896.00 | ||||||
| 1 | 1 | 1.00 | 11.00 | ||||||
| 2 | 1 | 2.00 | 16.00 | ||||||
| 4 | 1 | 4.00 | 176.00 | ||||||
| 8 | 1 | 8.00 | 3280.00 | ||||||
| 16 | 1 | 16.00 | 58256.00 |
Polynomial Roots Calculator found no rational roots
Polynomial Roots Calculator :
1.5 Find roots (zeroes) of : F(x) = x5-32
See theory in step 1.2
In this case, the Leading Coefficient is 1 and the Trailing Constant is -32.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,8 ,16 ,32
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -33.00 | ||||||
| -2 | 1 | -2.00 | -64.00 | ||||||
| -4 | 1 | -4.00 | -1056.00 | ||||||
| -8 | 1 | -8.00 | -32800.00 | ||||||
| -16 | 1 | -16.00 | -1048608.00 | ||||||
| -32 | 1 | -32.00 | -33554464.00 | ||||||
| 1 | 1 | 1.00 | -31.00 | ||||||
| 2 | 1 | 2.00 | 0.00 | x-2 | |||||
| 4 | 1 | 4.00 | 992.00 | ||||||
| 8 | 1 | 8.00 | 32736.00 | ||||||
| 16 | 1 | 16.00 | 1048544.00 | ||||||
| 32 | 1 | 32.00 | 33554400.00 |
The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms
In our case this means that
x5-32
can be divided with x-2
Polynomial Long Division :
1.6 Polynomial Long Division
Dividing : x5-32
("Dividend")
By : x-2 ("Divisor")
| dividend | x5 | - | 32 | ||||||||||
| - divisor | * x4 | x5 | - | 2x4 | |||||||||
| remainder | 2x4 | - | 32 | ||||||||||
| - divisor | * 2x3 | 2x4 | - | 4x3 | |||||||||
| remainder | 4x3 | - | 32 | ||||||||||
| - divisor | * 4x2 | 4x3 | - | 8x2 | |||||||||
| remainder | 8x2 | - | 32 | ||||||||||
| - divisor | * 8x1 | 8x2 | - | 16x | |||||||||
| remainder | 16x | - | 32 | ||||||||||
| - divisor | * 16x0 | 16x | - | 32 | |||||||||
| remainder | 0 |
Quotient : x4+2x3+4x2+8x+16 Remainder: 0
Polynomial Roots Calculator :
1.7 Find roots (zeroes) of : F(x) = x4+2x3+4x2+8x+16
See theory in step 1.2
In this case, the Leading Coefficient is 1 and the Trailing Constant is 16.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,8 ,16
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 11.00 | ||||||
| -2 | 1 | -2.00 | 16.00 | ||||||
| -4 | 1 | -4.00 | 176.00 | ||||||
| -8 | 1 | -8.00 | 3280.00 | ||||||
| -16 | 1 | -16.00 | 58256.00 | ||||||
| 1 | 1 | 1.00 | 31.00 | ||||||
| 2 | 1 | 2.00 | 80.00 | ||||||
| 4 | 1 | 4.00 | 496.00 | ||||||
| 8 | 1 | 8.00 | 5456.00 | ||||||
| 16 | 1 | 16.00 | 74896.00 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 1 :
(x4-2x3+4x2-8x+16)•(x+2)•(x4+2x3+4x2+8x+16)•(x-2) = 0
Step 2 :
Theory - Roots of a product :
2.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Quartic Equations :
2.2 Solve x4-2x3+4x2-8x+16 = 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
No interval at which a change of sign occures has been found. Consequently, Bisection Approximation can not be used. As this is a polynomial of an even degree it may not even have any real (as opposed to imaginary) roots
Solving a Single Variable Equation :
2.3 Solve : x+2 = 0
Subtract 2 from both sides of the equation :
x = -2
Quartic Equations :
2.4 Solve x4+2x3+4x2+8x+16 = 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
No interval at which a change of sign occures has been found. Consequently, Bisection Approximation can not be used. As this is a polynomial of an even degree it may not even have any real (as opposed to imaginary) roots
Solving a Single Variable Equation :
2.5 Solve : x-2 = 0
Add 2 to both sides of the equation :
x = 2
Two solutions were found :
- x = 2
- x = -2
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