Solution - Finding the roots of polynomials
Other Ways to Solve
Finding the roots of polynomialsStep by Step Solution
Step 1 :
Equation at the end of step 1 :
((((2•(x4))-(3•(x3)))-(2•32x2))+6x)+28Step 2 :
Equation at the end of step 2 :
((((2•(x4))-3x3)-(2•32x2))+6x)+28Step 3 :
Equation at the end of step 3 :
(((2x4 - 3x3) - (2•32x2)) + 6x) + 28
Step 4 :
Polynomial Roots Calculator :
4.1 Find roots (zeroes) of : F(x) = 2x4-3x3-18x2+6x+28
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 28.
The factor(s) are:
of the Leading Coefficient : 1,2
of the Trailing Constant : 1 ,2 ,4 ,7 ,14 ,28
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 9.00 | ||||||
| -1 | 2 | -0.50 | 21.00 | ||||||
| -2 | 1 | -2.00 | 0.00 | x+2 | |||||
| -4 | 1 | -4.00 | 420.00 | ||||||
| -7 | 1 | -7.00 | 4935.00 | ||||||
| -7 | 2 | -3.50 | 215.25 | ||||||
| -14 | 1 | -14.00 | 81480.00 | ||||||
| -28 | 1 | -28.00 | 1280916.00 | ||||||
| 1 | 1 | 1.00 | 15.00 | ||||||
| 1 | 2 | 0.50 | 26.25 | ||||||
| 2 | 1 | 2.00 | -24.00 | ||||||
| 4 | 1 | 4.00 | 84.00 | ||||||
| 7 | 1 | 7.00 | 2961.00 | ||||||
| 7 | 2 | 3.50 | 0.00 | 2x-7 | |||||
| 14 | 1 | 14.00 | 65184.00 | ||||||
| 28 | 1 | 28.00 | 1149540.00 |
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
2x4-3x3-18x2+6x+28
can be divided by 2 different polynomials,including by 2x-7
Polynomial Long Division :
4.2 Polynomial Long Division
Dividing : 2x4-3x3-18x2+6x+28
("Dividend")
By : 2x-7 ("Divisor")
| dividend | 2x4 | - | 3x3 | - | 18x2 | + | 6x | + | 28 | ||
| - divisor | * x3 | 2x4 | - | 7x3 | |||||||
| remainder | 4x3 | - | 18x2 | + | 6x | + | 28 | ||||
| - divisor | * 2x2 | 4x3 | - | 14x2 | |||||||
| remainder | - | 4x2 | + | 6x | + | 28 | |||||
| - divisor | * -2x1 | - | 4x2 | + | 14x | ||||||
| remainder | - | 8x | + | 28 | |||||||
| - divisor | * -4x0 | - | 8x | + | 28 | ||||||
| remainder | 0 |
Quotient : x3+2x2-2x-4 Remainder: 0
Polynomial Roots Calculator :
4.3 Find roots (zeroes) of : F(x) = x3+2x2-2x-4
See theory in step 4.1
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 | -1.00 | ||||||
| -2 | 1 | -2.00 | 0.00 | x+2 | |||||
| -4 | 1 | -4.00 | -28.00 | ||||||
| 1 | 1 | 1.00 | -3.00 | ||||||
| 2 | 1 | 2.00 | 8.00 | ||||||
| 4 | 1 | 4.00 | 84.00 |
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+2x2-2x-4
can be divided with x+2
Polynomial Long Division :
4.4 Polynomial Long Division
Dividing : x3+2x2-2x-4
("Dividend")
By : x+2 ("Divisor")
| dividend | x3 | + | 2x2 | - | 2x | - | 4 | ||
| - divisor | * x2 | x3 | + | 2x2 | |||||
| remainder | - | 2x | - | 4 | |||||
| - divisor | * 0x1 | ||||||||
| remainder | - | 2x | - | 4 | |||||
| - divisor | * -2x0 | - | 2x | - | 4 | ||||
| remainder | 0 |
Quotient : x2-2 Remainder: 0
Trying to factor as a Difference of Squares :
4.5 Factoring: x2-2
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 2 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares.
Final result :
(x2 - 2) • (x + 2) • (2x - 7)
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