Solution - Linear equations with one unknown
Other Ways to Solve
Linear equations with one unknownStep by Step Solution
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
f*(x)-((x^(2)*1)/(3-x^(3)-2*x))=0
Step by step solution :
Step 1 :
x2
Simplify ————————————
-x3 - 2x + 3
Step 2 :
Pulling out like terms :
2.1 Pull out like factors :
-x3 - 2x + 3 = -1 • (x3 + 2x - 3)
Polynomial Roots Calculator :
2.2 Find roots (zeroes) of : F(x) = x3 + 2x - 3
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 -3.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,3
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -6.00 | ||||||
| -3 | 1 | -3.00 | -36.00 | ||||||
| 1 | 1 | 1.00 | 0.00 | x - 1 | |||||
| 3 | 1 | 3.00 | 30.00 |
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 + 2x - 3
can be divided with x - 1
Polynomial Long Division :
2.3 Polynomial Long Division
Dividing : x3 + 2x - 3
("Dividend")
By : x - 1 ("Divisor")
| dividend | x3 | + | 2x | - | 3 | ||||
| - divisor | * x2 | x3 | - | x2 | |||||
| remainder | x2 | + | 2x | - | 3 | ||||
| - divisor | * x1 | x2 | - | x | |||||
| remainder | 3x | - | 3 | ||||||
| - divisor | * 3x0 | 3x | - | 3 | |||||
| remainder | 0 |
Quotient : x2+x+3 Remainder: 0
Trying to factor by splitting the middle term
2.4 Factoring x2+x+3
The first term is, x2 its coefficient is 1 .
The middle term is, +x its coefficient is 1 .
The last term, "the constant", is +3
Step-1 : Multiply the coefficient of the first term by the constant 1 • 3 = 3
Step-2 : Find two factors of 3 whose sum equals the coefficient of the middle term, which is 1 .
| -3 | + | -1 | = | -4 | ||
| -1 | + | -3 | = | -4 | ||
| 1 | + | 3 | = | 4 | ||
| 3 | + | 1 | = | 4 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 2 :
-x2
fx - —————————————————————— = 0
(x2 + x + 3) • (x - 1)
Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1 Subtracting a fraction from a whole
Rewrite the whole as a fraction using (x2+x+3) • (x-1) as the denominator :
fx fx • (x2 + x + 3) • (x - 1)
fx = —— = ———————————————————————————
1 (x2 + x + 3) • (x - 1)
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Adding fractions that have a common denominator :
3.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
fx • (x2+x+3) • (x-1) - (-x2) fx4 + 2fx2 - 3fx + x2
————————————————————————————— = ——————————————————————————
1 • (x2+x+3) • (x-1) 1 • (x2 + x + 3) • (x - 1)
Step 4 :
Pulling out like terms :
4.1 Pull out like factors :
fx4 + 2fx2 - 3fx + x2 =
x • (fx3 + 2fx - 3f + x)
Checking for a perfect cube :
4.2 fx3 + 2fx - 3f + x is not a perfect cube
Equation at the end of step 4 :
x • (fx3 + 2fx - 3f + x)
———————————————————————— = 0
(x2 + x + 3) • (x - 1)
Step 5 :
When a fraction equals zero :
5.1 When a fraction equals zero ...Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.
Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.
Here's how:
x•(fx3+2fx-3f+x)
———————————————— • (x2+x+3)•(x-1) = 0 • (x2+x+3)•(x-1)
(x2+x+3)•(x-1)
Now, on the left hand side, the (x2+x+3) • (x-1) cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
x • (fx3+2fx-3f+x) = 0
Theory - Roots of a product :
5.2 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 :
5.3 Solve : x = 0
Solution is x = 0
Solving a Single Variable Equation :
5.4 Solve fx3+2fx-3f+x = 0
In this type of equations, having more than one variable (unknown), you have to specify for which variable you want the equation solved.
We shall not handle this type of equations at this time.
One solution was found :
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