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Finding the roots of polynomials

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We think you wrote:

y^3+5y+6y^2+6(y+1)

This solution deals with finding the roots (zeroes) of polynomials.

Solution found

(y+2)*(y+1)*(y+3)
(y+2)*(y+1)*(y+3)

Step by Step Solution

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Step  1  :

Equation at the end of step  1  :

  (((y3)+5y)+(6•(y2)))+6•(y+1)

Step  2  :

Equation at the end of step  2  :

  (((y3) +  5y) +  (2•3y2)) +  6 • (y + 1)

Step  3  :

Checking for a perfect cube :

 3.1    y3+6y2+11y+6  is not a perfect cube

Trying to factor by pulling out :

 3.2      Factoring:  y3+6y2+11y+6 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  11y+6 
Group 2:  6y2+y3 

Pull out from each group separately :

Group 1:   (11y+6) • (1)
Group 2:   (y+6) • (y2)

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 :

 3.3    Find roots (zeroes) of :       F(y) = y3+6y2+11y+6
Polynomial Roots Calculator is a set of methods aimed at finding values of  y  for which   F(y)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  y  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  6.

 
The factor(s) are:

of the Leading Coefficient :  1
 
of the Trailing Constant :  1 ,2 ,3 ,6

 
Let us test ....

  P  Q  P/Q  F(P/Q)   Divisor
     -1     1      -1.00      0.00    y+1 
     -2     1      -2.00      0.00    y+2 
     -3     1      -3.00      0.00    y+3 
     -6     1      -6.00      -60.00   
     1     1      1.00      24.00   
     2     1      2.00      60.00   
     3     1      3.00      120.00   
     6     1      6.00      504.00   


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
   y3+6y2+11y+6 
can be divided by 3 different polynomials,including by  y+3 

Polynomial Long Division :

 3.4    Polynomial Long Division
Dividing :  y3+6y2+11y+6 
                              ("Dividend")
By         :    y+3    ("Divisor")

dividend  y3 + 6y2 + 11y + 6 
- divisor * y2   y3 + 3y2     
remainder    3y2 + 11y + 6 
- divisor * 3y1     3y2 + 9y   
remainder      2y + 6 
- divisor * 2y0       2y + 6 
remainder       0

Quotient :  y2+3y+2  Remainder:  0 

Trying to factor by splitting the middle term

 3.5     Factoring  y2+3y+2 

The first term is,  y2  its coefficient is  1 .
The middle term is,  +3y  its coefficient is  3 .
The last term, "the constant", is  +2 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 2 = 2 

Step-2 : Find two factors of  2  whose sum equals the coefficient of the middle term, which is   3 .

     -2   +   -1   =   -3
     -1   +   -2   =   -3
     1   +   2   =   3   That's it


Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  1  and  2 
                     y2 + 1y + 2y + 2

Step-4 : Add up the first 2 terms, pulling out like factors :
                    y • (y+1)
              Add up the last 2 terms, pulling out common factors :
                    2 • (y+1)
Step-5 : Add up the four terms of step 4 :
                    (y+2)  •  (y+1)
             Which is the desired factorization

Final result :

  (y + 2) • (y + 1) • (y + 3)

Why learn this

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