Solution - Finding the roots of polynomials
Other Ways to Solve
Finding the roots of polynomialsStep by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "x2" was replaced by "x^2". 2 more similar replacement(s).
Step 1 :
1
Simplify ——
x2
Equation at the end of step 1 :
1
((((x3)-(x2))+x)-——)-1
x2
Step 2 :
Rewriting the whole as an Equivalent Fraction :
2.1 Subtracting a fraction from a whole
Rewrite the whole as a fraction using x2 as the denominator :
x3 - x2 + x (x3 - x2 + x) • x2
x3 - x2 + x = ——————————— = ——————————————————
1 x2
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Step 3 :
Pulling out like terms :
3.1 Pull out like factors :
x3 - x2 + x = x • (x2 - x + 1)
Trying to factor by splitting the middle term
3.2 Factoring x2 - x + 1
The first term is, x2 its coefficient is 1 .
The middle term is, -x its coefficient is -1 .
The last term, "the constant", is +1
Step-1 : Multiply the coefficient of the first term by the constant 1 • 1 = 1
Step-2 : Find two factors of 1 whose sum equals the coefficient of the middle term, which is -1 .
| -1 | + | -1 | = | -2 | ||
| 1 | + | 1 | = | 2 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Adding fractions that have a common denominator :
3.3 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
x • (x2-x+1) • x2 - (1) x5 - x4 + x3 - 1
——————————————————————— = ————————————————
x2 x2
Equation at the end of step 3 :
(x5 - x4 + x3 - 1)
—————————————————— - 1
x2
Step 4 :
Rewriting the whole as an Equivalent Fraction :
4.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using x2 as the denominator :
1 1 • x2
1 = — = ——————
1 x2
Checking for a perfect cube :
4.2 x5 - x4 + x3 - 1 is not a perfect cube
Trying to factor by pulling out :
4.3 Factoring: x5 - x4 + x3 - 1
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: x3 - 1
Group 2: x5 - x4
Pull out from each group separately :
Group 1: (x3 - 1) • (1)
Group 2: (x - 1) • (x4)
Bad news !! Factoring by pulling out fails :
The groups have no common factor and can not be added up to form a multiplication.
Polynomial Roots Calculator :
4.4 Find roots (zeroes) of : F(x) = x5 - x4 + x3 - 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 | -4.00 | ||||||
| 1 | 1 | 1.00 | 0.00 | x - 1 |
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 - x4 + x3 - 1
can be divided with x - 1
Polynomial Long Division :
4.5 Polynomial Long Division
Dividing : x5 - x4 + x3 - 1
("Dividend")
By : x - 1 ("Divisor")
| dividend | x5 | - | x4 | + | x3 | - | 1 | ||||||
| - divisor | * x4 | x5 | - | x4 | |||||||||
| remainder | x3 | - | 1 | ||||||||||
| - divisor | * 0x3 | ||||||||||||
| remainder | x3 | - | 1 | ||||||||||
| - divisor | * x2 | x3 | - | x2 | |||||||||
| remainder | x2 | - | 1 | ||||||||||
| - divisor | * x1 | x2 | - | x | |||||||||
| remainder | x | - | 1 | ||||||||||
| - divisor | * x0 | x | - | 1 | |||||||||
| remainder | 0 |
Quotient : x4+x2+x+1 Remainder: 0
Polynomial Roots Calculator :
4.6 Find roots (zeroes) of : F(x) = x4+x2+x+1
See theory in step 4.4
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 | 2.00 | ||||||
| 1 | 1 | 1.00 | 4.00 |
Polynomial Roots Calculator found no rational roots
Adding fractions that have a common denominator :
4.7 Adding up the two equivalent fractions
(x4+x2+x+1) • (x-1) - (x2) x5 - x4 + x3 - x2 - 1
—————————————————————————— = —————————————————————
x2 x2
Polynomial Roots Calculator :
4.8 Find roots (zeroes) of : F(x) = x5 - x4 + x3 - x2 - 1
See theory in step 4.4
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 | -1.00 |
Polynomial Roots Calculator found no rational roots
Final result :
x5 - x4 + x3 - x2 - 1
—————————————————————
x2
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