Solution - Finding the roots of polynomials
Other Ways to Solve
Finding the roots of polynomialsStep by Step Solution
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((((2•(x4))-(x3))+17x2)-9x)-9 = 0Step 2 :
Equation at the end of step 2 :
(((2x4 - x3) + 17x2) - 9x) - 9 = 0
Step 3 :
Polynomial Roots Calculator :
3.1 Find roots (zeroes) of : F(x) = 2x4-x3+17x2-9x-9
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 2 and the Trailing Constant is -9.
The factor(s) are:
of the Leading Coefficient : 1,2
of the Trailing Constant : 1 ,3 ,9
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 20.00 | ||||||
| -1 | 2 | -0.50 | 0.00 | 2x+1 | |||||
| -3 | 1 | -3.00 | 360.00 | ||||||
| -3 | 2 | -1.50 | 56.25 | ||||||
| -9 | 1 | -9.00 | 15300.00 | ||||||
| -9 | 2 | -4.50 | 1287.00 | ||||||
| 1 | 1 | 1.00 | 0.00 | x-1 | |||||
| 1 | 2 | 0.50 | -9.25 | ||||||
| 3 | 1 | 3.00 | 252.00 | ||||||
| 3 | 2 | 1.50 | 22.50 | ||||||
| 9 | 1 | 9.00 | 13680.00 | ||||||
| 9 | 2 | 4.50 | 1023.75 |
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
2x4-x3+17x2-9x-9
can be divided by 2 different polynomials,including by x-1
Polynomial Long Division :
3.2 Polynomial Long Division
Dividing : 2x4-x3+17x2-9x-9
("Dividend")
By : x-1 ("Divisor")
| dividend | 2x4 | - | x3 | + | 17x2 | - | 9x | - | 9 | ||
| - divisor | * 2x3 | 2x4 | - | 2x3 | |||||||
| remainder | x3 | + | 17x2 | - | 9x | - | 9 | ||||
| - divisor | * x2 | x3 | - | x2 | |||||||
| remainder | 18x2 | - | 9x | - | 9 | ||||||
| - divisor | * 18x1 | 18x2 | - | 18x | |||||||
| remainder | 9x | - | 9 | ||||||||
| - divisor | * 9x0 | 9x | - | 9 | |||||||
| remainder | 0 |
Quotient : 2x3+x2+18x+9 Remainder: 0
Polynomial Roots Calculator :
3.3 Find roots (zeroes) of : F(x) = 2x3+x2+18x+9
See theory in step 3.1
In this case, the Leading Coefficient is 2 and the Trailing Constant is 9.
The factor(s) are:
of the Leading Coefficient : 1,2
of the Trailing Constant : 1 ,3 ,9
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -10.00 | ||||||
| -1 | 2 | -0.50 | 0.00 | 2x+1 | |||||
| -3 | 1 | -3.00 | -90.00 | ||||||
| -3 | 2 | -1.50 | -22.50 | ||||||
| -9 | 1 | -9.00 | -1530.00 | ||||||
| -9 | 2 | -4.50 | -234.00 | ||||||
| 1 | 1 | 1.00 | 30.00 | ||||||
| 1 | 2 | 0.50 | 18.50 | ||||||
| 3 | 1 | 3.00 | 126.00 | ||||||
| 3 | 2 | 1.50 | 45.00 | ||||||
| 9 | 1 | 9.00 | 1710.00 | ||||||
| 9 | 2 | 4.50 | 292.50 |
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
2x3+x2+18x+9
can be divided with 2x+1
Polynomial Long Division :
3.4 Polynomial Long Division
Dividing : 2x3+x2+18x+9
("Dividend")
By : 2x+1 ("Divisor")
| dividend | 2x3 | + | x2 | + | 18x | + | 9 | ||
| - divisor | * x2 | 2x3 | + | x2 | |||||
| remainder | 18x | + | 9 | ||||||
| - divisor | * 0x1 | ||||||||
| remainder | 18x | + | 9 | ||||||
| - divisor | * 9x0 | 18x | + | 9 | |||||
| remainder | 0 |
Quotient : x2+9 Remainder: 0
Polynomial Roots Calculator :
3.5 Find roots (zeroes) of : F(x) = x2+9
See theory in step 3.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is 9.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,3 ,9
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 10.00 | ||||||
| -3 | 1 | -3.00 | 18.00 | ||||||
| -9 | 1 | -9.00 | 90.00 | ||||||
| 1 | 1 | 1.00 | 10.00 | ||||||
| 3 | 1 | 3.00 | 18.00 | ||||||
| 9 | 1 | 9.00 | 90.00 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 3 :
(x2 + 9) • (2x + 1) • (x - 1) = 0
Step 4 :
Theory - Roots of a product :
4.1 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.
Solving a Single Variable Equation :
4.2 Solve : x2+9 = 0
Subtract 9 from both sides of the equation :
x2 = -9
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ -9
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Accordingly, √ -9 =
√ -1• 9 =
√ -1 •√ 9 =
i • √ 9
Can √ 9 be simplified ?
Yes! The prime factorization of 9 is
3•3
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 9 = √ 3•3 =
± 3 • √ 1 =
± 3
The equation has no real solutions. It has 2 imaginary, or complex solutions.
x= 0.0000 + 3.0000 i
x= 0.0000 - 3.0000 i
Solving a Single Variable Equation :
4.3 Solve : 2x+1 = 0
Subtract 1 from both sides of the equation :
2x = -1
Divide both sides of the equation by 2:
x = -1/2 = -0.500
Solving a Single Variable Equation :
4.4 Solve : x-1 = 0
Add 1 to both sides of the equation :
x = 1
Four solutions were found :
- x = 1
- x = -1/2 = -0.500
- x= 0.0000 - 3.0000 i
- x= 0.0000 + 3.0000 i
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