Solve Polynomialrootcalculator (x^4-5x)^2-36=0 Tiger Algebra Solver

Step by step solution :

Step 1 :

1.1 Evaluate : (x4-5x)2 = x8-10x5+25x2

Checking for a perfect cube :

1.2 x8-10x5+25x2-36 is not a perfect cube

Trying to factor by pulling out :

1.3 Factoring: x8-10x5+25x2-36 Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: 25x2-36 Group 2: -10x5+x8 Pull out from each group separately :Group 1: (25x2-36) • (1)Group 2: (x3-10) • (x5)Bad news !! Factoring by pulling out fails : The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

1.4 Find roots (zeroes) of : F(x) = x8-10x5+25x2-36Polynomial 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 integersThe 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 CoefficientIn this case, the Leading Coefficient is 1 and the Trailing Constant is -36. The factor(s) are: of the Leading Coefficient : 1 of the Trailing Constant : 1 ,2 ,3 ,4 ,6 ,9 ,12 ,18 ,36 Let us test ....

P	Q	P/Q	F(P/Q)	Divisor
-1	1	-1.00	0.00	x+1
-2	1	-2.00	640.00
-3	1	-3.00	9180.00
-4	1	-4.00	76140.00
-6	1	-6.00	1758240.00
-9	1	-9.00	43639200.00
-12	1	-12.00	432473580.00
-18	1	-18.00	11038864320.00
-36	1	-36.00	2821714601580.00
1	1	1.00	-20.00
2	1	2.00	0.00	x-2
3	1	3.00	4320.00
4	1	4.00	55660.00
6	1	6.00	1602720.00
9	1	9.00	42458220.00
12	1	12.00	427496940.00
18	1	18.00	11001072960.00
36	1	36.00	2820505278060.00

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 x8-10x5+25x2-36 can be divided by 2 different polynomials,including by x-2

Polynomial Long Division :

1.5 Polynomial Long Division Dividing : x8-10x5+25x2-36 ("Dividend") By : x-2 ("Divisor")

dividend	x8	-	10x5	+	25x2	-	36
- divisor	* x7	x8	-	2x7
remainder	2x7	-	10x5	+	25x2	-	36
- divisor	* 2x6	2x7	-	4x6
remainder	4x6	-	10x5	+	25x2	-	36
- divisor	* 4x5	4x6	-	8x5
remainder	-	2x5	+	25x2	-	36
- divisor	* -2x4	-	2x5	+	4x4
remainder	-	4x4	+	25x2	-	36
- divisor	* -4x3	-	4x4	+	8x3
remainder	-	8x3	+	25x2	-	36
- divisor	* -8x2	-	8x3	+	16x2
remainder	9x2	-	36
- divisor	* 9x1	9x2	-	18x
remainder	18x	-	36
- divisor	* 18x0	18x	-	36
remainder	0

Quotient : x7+2x6+4x5-2x4-4x3-8x2+9x+18 Remainder: 0

Equation at the end of step 1 :

(x7 + 2x6 + 4x5 - 2x4 - 4x3 - 8x2 + 9x + 18) • (x - 2) = 0

Step 2 :

Theory - Roots of a product :

2.1 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.

Equations of order 5 or higher :

2.2 Solve x7+2x6+4x5-2x4-4x3-8x2+9x+18 = 0Points regarding equations of degree five or higher. (1) There is no general method (Formula) for solving polynomial equations of degree five or higher. (2) By the Fundamental theorem of Algebra, if we allow complex numbers, an equation of degree n will have exactly n solutions (This is if we count double solutions as 2 , triple solutions as 3 and so on) (3) By the Abel-Ruffini theorem, the solutions can not always be presented in the conventional way using only a finite amount of additions, subtractions, multiplications, divisions or root extractions

(4) If F(x) is a polynomial of odd degree with real coefficients, then the equation F(X)=0 has at least one real solution. (5) Using methods such as the Bisection Method, real solutions can be approximated to any desired degree of accuracy. Failed to find the initial interval for implementing the BiSection Method

Solving a Single Variable Equation :

2.3 Solve : x-2 = 0Add 2 to both sides of the equation : x = 2

One solution was found :

x = 2