Solution - Finding the roots of polynomials
Step by Step Solution
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((((x4)-(x3))+2x2)-4x)-8 = 0
Step 2 :
Polynomial Roots Calculator :
2.1 Find roots (zeroes) of : F(x) = x4-x3+2x2-4x-8
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 -8.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,8
Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | 0.00 | x+1 | |||||
-2 | 1 | -2.00 | 32.00 | ||||||
-4 | 1 | -4.00 | 360.00 | ||||||
-8 | 1 | -8.00 | 4760.00 | ||||||
1 | 1 | 1.00 | -10.00 | ||||||
2 | 1 | 2.00 | 0.00 | x-2 | |||||
4 | 1 | 4.00 | 200.00 | ||||||
8 | 1 | 8.00 | 3672.00 |
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-x3+2x2-4x-8
can be divided by 2 different polynomials,including by x-2
Polynomial Long Division :
2.2 Polynomial Long Division
Dividing : x4-x3+2x2-4x-8
("Dividend")
By : x-2 ("Divisor")
dividend | x4 | - | x3 | + | 2x2 | - | 4x | - | 8 | ||
- divisor | * x3 | x4 | - | 2x3 | |||||||
remainder | x3 | + | 2x2 | - | 4x | - | 8 | ||||
- divisor | * x2 | x3 | - | 2x2 | |||||||
remainder | 4x2 | - | 4x | - | 8 | ||||||
- divisor | * 4x1 | 4x2 | - | 8x | |||||||
remainder | 4x | - | 8 | ||||||||
- divisor | * 4x0 | 4x | - | 8 | |||||||
remainder | 0 |
Quotient : x3+x2+4x+4 Remainder: 0
Polynomial Roots Calculator :
2.3 Find roots (zeroes) of : F(x) = x3+x2+4x+4
See theory in step 2.1
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 | 0.00 | x+1 | |||||
-2 | 1 | -2.00 | -8.00 | ||||||
-4 | 1 | -4.00 | -60.00 | ||||||
1 | 1 | 1.00 | 10.00 | ||||||
2 | 1 | 2.00 | 24.00 | ||||||
4 | 1 | 4.00 | 100.00 |
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 :
2.4 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 :
2.5 Find roots (zeroes) of : F(x) = x2+4
See theory in step 2.1
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 2 :
(x2 + 4) • (x + 1) • (x - 2) = 0
Step 3 :
Theory - Roots of a product :
3.1 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 :
3.2 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 :
3.3 Solve : x+1 = 0
Subtract 1 from both sides of the equation :
x = -1
Solving a Single Variable Equation :
3.4 Solve : x-2 = 0
Add 2 to both sides of the equation :
x = 2
Four solutions were found :
- x = 2
- x = -1
- x= 0.0000 - 2.0000 i
- x= 0.0000 + 2.0000 i
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