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 :
((((x4)-(4•(x3)))+7x2)-16x)+12 = 0Step 2 :
Equation at the end of step 2 :
((((x4) - 22x3) + 7x2) - 16x) + 12 = 0
Step 3 :
Polynomial Roots Calculator :
3.1 Find roots (zeroes) of : F(x) = x4-4x3+7x2-16x+12
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 12.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,4 ,6 ,12
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 40.00 | ||||||
| -2 | 1 | -2.00 | 120.00 | ||||||
| -3 | 1 | -3.00 | 312.00 | ||||||
| -4 | 1 | -4.00 | 700.00 | ||||||
| -6 | 1 | -6.00 | 2520.00 | ||||||
| -12 | 1 | -12.00 | 28860.00 | ||||||
| 1 | 1 | 1.00 | 0.00 | x-1 | |||||
| 2 | 1 | 2.00 | -8.00 | ||||||
| 3 | 1 | 3.00 | 0.00 | x-3 | |||||
| 4 | 1 | 4.00 | 60.00 | ||||||
| 6 | 1 | 6.00 | 600.00 | ||||||
| 12 | 1 | 12.00 | 14652.00 |
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
x4-4x3+7x2-16x+12
can be divided by 2 different polynomials,including by x-3
Polynomial Long Division :
3.2 Polynomial Long Division
Dividing : x4-4x3+7x2-16x+12
("Dividend")
By : x-3 ("Divisor")
| dividend | x4 | - | 4x3 | + | 7x2 | - | 16x | + | 12 | ||
| - divisor | * x3 | x4 | - | 3x3 | |||||||
| remainder | - | x3 | + | 7x2 | - | 16x | + | 12 | |||
| - divisor | * -x2 | - | x3 | + | 3x2 | ||||||
| remainder | 4x2 | - | 16x | + | 12 | ||||||
| - divisor | * 4x1 | 4x2 | - | 12x | |||||||
| remainder | - | 4x | + | 12 | |||||||
| - divisor | * -4x0 | - | 4x | + | 12 | ||||||
| remainder | 0 |
Quotient : x3-x2+4x-4 Remainder: 0
Polynomial Roots Calculator :
3.3 Find roots (zeroes) of : F(x) = x3-x2+4x-4
See theory in step 3.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is -4.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -10.00 | ||||||
| -2 | 1 | -2.00 | -24.00 | ||||||
| -4 | 1 | -4.00 | -100.00 | ||||||
| 1 | 1 | 1.00 | 0.00 | x-1 | |||||
| 2 | 1 | 2.00 | 8.00 | ||||||
| 4 | 1 | 4.00 | 60.00 |
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
x3-x2+4x-4
can be divided with x-1
Polynomial Long Division :
3.4 Polynomial Long Division
Dividing : x3-x2+4x-4
("Dividend")
By : x-1 ("Divisor")
| dividend | x3 | - | x2 | + | 4x | - | 4 | ||
| - divisor | * x2 | x3 | - | x2 | |||||
| remainder | 4x | - | 4 | ||||||
| - divisor | * 0x1 | ||||||||
| remainder | 4x | - | 4 | ||||||
| - divisor | * 4x0 | 4x | - | 4 | |||||
| remainder | 0 |
Quotient : x2+4 Remainder: 0
Polynomial Roots Calculator :
3.5 Find roots (zeroes) of : F(x) = x2+4
See theory in step 3.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is 4.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 5.00 | ||||||
| -2 | 1 | -2.00 | 8.00 | ||||||
| -4 | 1 | -4.00 | 20.00 | ||||||
| 1 | 1 | 1.00 | 5.00 | ||||||
| 2 | 1 | 2.00 | 8.00 | ||||||
| 4 | 1 | 4.00 | 20.00 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 3 :
(x2 + 4) • (x - 1) • (x - 3) = 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+4 = 0
Subtract 4 from both sides of the equation :
x2 = -4
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ -4
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Accordingly, √ -4 =
√ -1• 4 =
√ -1 •√ 4 =
i • √ 4
Can √ 4 be simplified ?
Yes! The prime factorization of 4 is
2•2
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).
√ 4 = √ 2•2 =
± 2 • √ 1 =
± 2
The equation has no real solutions. It has 2 imaginary, or complex solutions.
x= 0.0000 + 2.0000 i
x= 0.0000 - 2.0000 i
Solving a Single Variable Equation :
4.3 Solve : x-1 = 0
Add 1 to both sides of the equation :
x = 1
Solving a Single Variable Equation :
4.4 Solve : x-3 = 0
Add 3 to both sides of the equation :
x = 3
Four solutions were found :
- x = 3
- x = 1
- x= 0.0000 - 2.0000 i
- x= 0.0000 + 2.0000 i
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