Question

x3 - 6x2 + 11x - 6 / x-3

Answer 1

 (x3-6x2+11x-6):(x-3) Final result : (x - 1) • (x - 2) Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x2"   was replaced by   "x^2".  1 more similar replacement(s).

Step by step solution :Step  1  :Equation at the end of step  1  : Step  2  : x3 - 6x2 + 11x - 6 Simplify —————————————————— x - 3 Checking for a perfect cube :

 2.1    x3 - 6x2 + 11x - 6  is not a perfect cube 

Trying to factor by pulling out :

 2.2      Factoring:  x3 - 6x2 + 11x - 6 

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

Group 1:  11x - 6 
Group 2:  -6x2 + x3 

Pull out from each group separately :

Group 1:   (11x - 6) • (1)
Group 2:   (x - 6) • (x2)

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 :

 2.3    Find roots (zeroes) of :       F(x) = x3 - 6x2 + 11x - 6
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  -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      -24.00        -2     1      -2.00      -60.00        -3     1      -3.00      -120.00        -6     1      -6.00      -504.00        1     1      1.00      0.00    x - 1      2     1      2.00      0.00    x - 2      3     1      3.00      0.00    x - 3      6     1      6.00      60.00   

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 - 6x2 + 11x - 6 
can be divided by 3 different polynomials,including by  x - 3 

Polynomial Long Division :

 2.4    Polynomial Long Division 
Dividing :  x3 - 6x2 + 11x - 6 
                              ("Dividend")
By         :    x - 3    ("Divisor")

dividend  x3 - 6x2 + 11x - 6 - divisor * x2   x3 - 3x2     remainder  - 3x2 + 11x - 6 - divisor * -3x1   - 3x2 + 9x   remainder      2x - 6 - divisor * 2x0       2x - 6 remainder       0

Quotient :  x2-3x+2  Remainder:  0 

Trying to factor by splitting the middle term

 2.5     Factoring  x2-3x+2 

The first term is,  x2  its coefficient is  1 .
The middle term is,  -3x  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   That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -2  and  -1 
                     x2 - 2x - 1x - 2

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

Canceling Out :

 2.6    Cancel out  (x-3)  which appears on both sides of the fraction line.

Final result : (x - 1) • (x - 2)