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Changes made to your input should not affect the solution:
(1): "x2" was replaced by "x^2". 2 more similar replacement(s).
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((((x4)+(x3))-23x2)-3x)+60 = 0Step 2 :
Polynomial Roots Calculator :
2.1 Find roots (zeroes) of : F(x) = x4+x3-23x2-3x+60
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 The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant :
1 ,2 ,3 ,4 ,5 ,6 ,10 ,12 ,15 ,20 , etc
Let us test ....-1 | 1 | -1.00 | 40.00 | ||||||
-2 | 1 | -2.00 | -18.00 | ||||||
-3 | 1 | -3.00 | -84.00 | ||||||
-4 | 1 | -4.00 | -104.00 | ||||||
-5 | 1 | -5.00 | 0.00 | x+5 | |||||
-6 | 1 | -6.00 | 330.00 | ||||||
-10 | 1 | -10.00 | 6790.00 | ||||||
-12 | 1 | -12.00 | 15792.00 | ||||||
-15 | 1 | -15.00 | 42180.00 | ||||||
-20 | 1 | -20.00 | 142920.00 | ||||||
1 | 1 | 1.00 | 36.00 | ||||||
2 | 1 | 2.00 | -14.00 | ||||||
3 | 1 | 3.00 | -48.00 | ||||||
4 | 1 | 4.00 | 0.00 | x-4 | |||||
5 | 1 | 5.00 | 220.00 | ||||||
6 | 1 | 6.00 | 726.00 | ||||||
10 | 1 | 10.00 | 8730.00 | ||||||
12 | 1 | 12.00 | 19176.00 | ||||||
15 | 1 | 15.00 | 48840.00 | ||||||
20 | 1 | 20.00 | 158800.00 |
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+x3-23x2-3x+60
can be divided by 2 different polynomials,including by x-4
Polynomial Long Division :
2.2 Polynomial Long Division
Dividing :
x4+x3-23x2-3x+60 ("Dividend")
By : x-4 ("Divisor")dividend | x4 | + | x3 | - | 23x2 | - | 3x | + | 60 | ||
- divisor | * x3 | x4 | - | 4x3 | |||||||
remainder | 5x3 | - | 23x2 | - | 3x | + | 60 | ||||
- divisor | * 5x2 | 5x3 | - | 20x2 | |||||||
remainder | - | 3x2 | - | 3x | + | 60 | |||||
- divisor | * -3x1 | - | 3x2 | + | 12x | ||||||
remainder | - | 15x | + | 60 | |||||||
- divisor | * -15x0 | - | 15x | + | 60 | ||||||
remainder | 0 |
Quotient : x3+5x2-3x-15 Remainder: 0
Polynomial Roots Calculator :
2.3 Find roots (zeroes) of : F(x) = x3+5x2-3x-15
See theory in step 2.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : Let us test ....
-1 | 1 | -1.00 | -8.00 | ||||||
-3 | 1 | -3.00 | 12.00 | ||||||
-5 | 1 | -5.00 | 0.00 | x+5 | |||||
-15 | 1 | -15.00 | -2220.00 | ||||||
1 | 1 | 1.00 | -12.00 | ||||||
3 | 1 | 3.00 | 48.00 | ||||||
5 | 1 | 5.00 | 220.00 | ||||||
15 | 1 | 15.00 | 4440.00 |
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+5x2-3x-15
can be divided with x+5
Polynomial Long Division :
2.4 Polynomial Long Division
Dividing :
x3+5x2-3x-15 ("Dividend")
By : x+5 ("Divisor")dividend | x3 | + | 5x2 | - | 3x | - | 15 | ||
- divisor | * x2 | x3 | + | 5x2 | |||||
remainder | - | 3x | - | 15 | |||||
- divisor | * 0x1 | ||||||||
remainder | - | 3x | - | 15 | |||||
- divisor | * -3x0 | - | 3x | - | 15 | ||||
remainder | 0 |
Quotient : x2-3 Remainder: 0
Trying to factor as a Difference of Squares :
2.5 Factoring: x2-3
Theory : A difference of two perfect squares, A2 - B2 can be factored into Proof :
(A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 3 is not a square !!Ruling : Binomial can not be factored as the difference of two perfect squares.
Equation at the end of step 2 :
(x2 - 3) • (x + 5) • (x - 4) = 0Step 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 separatelyIn other words, we are going to solve as many equations as there are terms in the productAny solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
3.2 Solve : x2-3 = 0Add 3 to both sides of the equation :
x2 = 3
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ 3 The equation has two real solutions
These solutions are x = ± √3 = ± 1.7321
Solving a Single Variable Equation :
3.3 Solve : x+5 = 0Subtract 5 from both sides of the equation :
x = -5
Solving a Single Variable Equation :
3.4 Solve : x-4 = 0Add 4 to both sides of the equation :
x = 4
Four solutions were found :
- x = 4
- x = -5
- x = ± √3 = ± 1.7321