Solution - Polynomial long division

Step  1  :

Equation at the end of step  1  :

  ((((x4)-(x3))-3x2)+5x)-2

Step  2  :

Polynomial Roots Calculator :

 2.1    Find roots (zeroes) of :       F(x) = x⁴-x³-3x²+5x-2
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  -2.

 The factor(s) are:

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

 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-x³-3x²+5x-2 
can be divided by 2 different polynomials,including by  x-1 

Polynomial Long Division :

 2.2    Polynomial Long Division
Dividing :  x⁴-x³-3x²+5x-2 
                              ("Dividend")
By         :    x-1    ("Divisor")

Quotient :  x³-3x+2  Remainder:  0 

Polynomial Roots Calculator :

 2.3    Find roots (zeroes) of :       F(x) = x³-3x+2

     See theory in step 2.1
In this case, the Leading Coefficient is  1  and the Trailing Constant is  2.

 The factor(s) are:

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

 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
   x³-3x+2 
can be divided by 2 different polynomials,including by  x-1 

Polynomial Long Division :

 2.4    Polynomial Long Division
Dividing :  x³-3x+2 
                              ("Dividend")
By         :    x-1    ("Divisor")

Quotient :  x²+x-2  Remainder:  0 

Trying to factor by splitting the middle term

 2.5     Factoring  x²+x-2 

The first term is,  x²  its coefficient is  1 .
The middle term is,  +x  its coefficient is  1 .
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   1 .

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

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

Multiplying Exponential Expressions :

 2.6    Multiply  (x-1)  by  (x-1) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (x-1)  and the exponents are :
          1 , as  (x-1)  is the same number as  (x-1)¹ 
 and   1 , as  (x-1)  is the same number as  (x-1)¹ 
The product is therefore,  (x-1)⁽¹⁺¹⁾ = (x-1)² 

Multiplying Exponential Expressions :

 2.7    Multiply  (x-1)²   by  (x-1) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (x-1)  and the exponents are :
          2
 and   1 , as  (x-1)  is the same number as  (x-1)¹ 
The product is therefore,  (x-1)⁽²⁺¹⁾ = (x-1)³ 

Final result :

  (x + 2) • (x - 1)3