Obtain all the zeros of the polynomial f x x 4 x 3 23 x 2 3x 60 if two of the zeros are √ 3 and √ 3

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  :

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

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 

Polynomial Long Division :

 2.2    Polynomial Long Division
Dividing :

 x4+x3-23x2-3x+60                               ("Dividend")

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

of the Leading Coefficient :  1
 of the Trailing Constant : 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+5x2-3x-15 

Polynomial Long Division :

 2.4    Polynomial Long Division
Dividing :

 x3+5x2-3x-15                               ("Dividend")

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.

Ruling : Binomial can not be factored as the difference of two perfect squares.

Equation at the end of step  2  :

Step  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 :

1.  x = 4
2.  x = -5
3.  x = ± √3 = ± 1.7321