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)-(3•(x3)))+(3•7x2))-75x)-100 = 0Step 2 :
Equation at the end of step 2 :
((((x4) - 3x3) + (3•7x2)) - 75x) - 100 = 0
Step 3 :
Polynomial Roots Calculator :
3.1 Find roots (zeroes) of : F(x) = x4-3x3+21x2-75x-100
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 -100.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,5 ,10 ,20 ,25 ,50 ,100
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 0.00 | x+1 | |||||
| -2 | 1 | -2.00 | 174.00 | ||||||
| -4 | 1 | -4.00 | 984.00 | ||||||
| -5 | 1 | -5.00 | 1800.00 | ||||||
| -10 | 1 | -10.00 | 15750.00 | ||||||
| -20 | 1 | -20.00 | 193800.00 | ||||||
| -25 | 1 | -25.00 | 452400.00 | ||||||
| -50 | 1 | -50.00 | 6681150.00 | ||||||
| -100 | 1 | -100.00 | 103217400.00 | ||||||
| 1 | 1 | 1.00 | -156.00 | ||||||
| 2 | 1 | 2.00 | -174.00 | ||||||
| 4 | 1 | 4.00 | 0.00 | x-4 | |||||
| 5 | 1 | 5.00 | 300.00 | ||||||
| 10 | 1 | 10.00 | 8250.00 | ||||||
| 20 | 1 | 20.00 | 142800.00 | ||||||
| 25 | 1 | 25.00 | 354900.00 | ||||||
| 50 | 1 | 50.00 | 5923650.00 | ||||||
| 100 | 1 | 100.00 | 97202400.00 |
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
x4-3x3+21x2-75x-100
can be divided by 2 different polynomials,including by x-4
Polynomial Long Division :
3.2 Polynomial Long Division
Dividing : x4-3x3+21x2-75x-100
("Dividend")
By : x-4 ("Divisor")
| dividend | x4 | - | 3x3 | + | 21x2 | - | 75x | - | 100 | ||
| - divisor | * x3 | x4 | - | 4x3 | |||||||
| remainder | x3 | + | 21x2 | - | 75x | - | 100 | ||||
| - divisor | * x2 | x3 | - | 4x2 | |||||||
| remainder | 25x2 | - | 75x | - | 100 | ||||||
| - divisor | * 25x1 | 25x2 | - | 100x | |||||||
| remainder | 25x | - | 100 | ||||||||
| - divisor | * 25x0 | 25x | - | 100 | |||||||
| remainder | 0 |
Quotient : x3+x2+25x+25 Remainder: 0
Polynomial Roots Calculator :
3.3 Find roots (zeroes) of : F(x) = x3+x2+25x+25
See theory in step 3.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is 25.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,5 ,25
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 0.00 | x+1 | |||||
| -5 | 1 | -5.00 | -200.00 | ||||||
| -25 | 1 | -25.00 | -15600.00 | ||||||
| 1 | 1 | 1.00 | 52.00 | ||||||
| 5 | 1 | 5.00 | 300.00 | ||||||
| 25 | 1 | 25.00 | 16900.00 |
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+25x+25
can be divided with x+1
Polynomial Long Division :
3.4 Polynomial Long Division
Dividing : x3+x2+25x+25
("Dividend")
By : x+1 ("Divisor")
| dividend | x3 | + | x2 | + | 25x | + | 25 | ||
| - divisor | * x2 | x3 | + | x2 | |||||
| remainder | 25x | + | 25 | ||||||
| - divisor | * 0x1 | ||||||||
| remainder | 25x | + | 25 | ||||||
| - divisor | * 25x0 | 25x | + | 25 | |||||
| remainder | 0 |
Quotient : x2+25 Remainder: 0
Polynomial Roots Calculator :
3.5 Find roots (zeroes) of : F(x) = x2+25
See theory in step 3.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is 25.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,5 ,25
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 26.00 | ||||||
| -5 | 1 | -5.00 | 50.00 | ||||||
| -25 | 1 | -25.00 | 650.00 | ||||||
| 1 | 1 | 1.00 | 26.00 | ||||||
| 5 | 1 | 5.00 | 50.00 | ||||||
| 25 | 1 | 25.00 | 650.00 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 3 :
(x2 + 25) • (x + 1) • (x - 4) = 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+25 = 0
Subtract 25 from both sides of the equation :
x2 = -25
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ -25
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Accordingly, √ -25 =
√ -1• 25 =
√ -1 •√ 25 =
i • √ 25
Can √ 25 be simplified ?
Yes! The prime factorization of 25 is
5•5
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).
√ 25 = √ 5•5 =
± 5 • √ 1 =
± 5
The equation has no real solutions. It has 2 imaginary, or complex solutions.
x= 0.0000 + 5.0000 i
x= 0.0000 - 5.0000 i
Solving a Single Variable Equation :
4.3 Solve : x+1 = 0
Subtract 1 from both sides of the equation :
x = -1
Solving a Single Variable Equation :
4.4 Solve : x-4 = 0
Add 4 to both sides of the equation :
x = 4
Four solutions were found :
- x = 4
- x = -1
- x= 0.0000 - 5.0000 i
- x= 0.0000 + 5.0000 i
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