Finding The Roots Of Polynomials - Tiger Algebra

Reformatting the input :

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 :

Step 2 :

Equation at the end of step 2 :

Step 3 :

Polynomial Roots Calculator :

3.1 Find roots (zeroes) of : F(x) = x4+2x3-5x2-4x+6Polynomial 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 6. The factor(s) are: of the Leading Coefficient : 1 of the Trailing Constant : 1 ,2 ,3 ,6 Let us test ....

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+2x3-5x2-4x+6 can be divided by 2 different polynomials,including by x-1

Polynomial Long Division :

3.2 Polynomial Long Division Dividing : x4+2x3-5x2-4x+6 ("Dividend") By : x-1 ("Divisor")

Quotient : x3+3x2-2x-6 Remainder: 0

Polynomial Roots Calculator :

3.3 Find roots (zeroes) of : F(x) = x3+3x2-2x-6 See theory in step 3.1 In this case, the Leading Coefficient is 1 and the Trailing Constant is -6. The factor(s) are: of the Leading Coefficient : 1 of the Trailing Constant : 1 ,2 ,3 ,6 Let us test ....

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+3x2-2x-6 can be divided with x+3

Polynomial Long Division :

3.4 Polynomial Long Division Dividing : x3+3x2-2x-6 ("Dividend") By : x+3 ("Divisor")

Quotient : x2-2 Remainder: 0

Trying to factor as a Difference of Squares :

3.5 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 - B2Note : 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.

Equation at the end of step 3 :

Step 4 :

Theory - Roots of a product :

4.1 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 :

4.2 Solve : x2-2 = 0Add 2 to both sides of the equation : x2 = 2 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get: x = ± √ 2 The equation has two real solutions These solutions are x = ± √2 = ± 1.4142

Solving a Single Variable Equation :

4.3 Solve : x+3 = 0Subtract 3 from both sides of the equation : x = -3

Solving a Single Variable Equation :

4.4 Solve : x-1 = 0Add 1 to both sides of the equation : x = 1

Four solutions were found :

1. x = 1
2. x = -3
3. x = ± √2 = ± 1.4142