Solution - Adding, subtracting and finding the least common multiple

Step  1  :

               2 - t   
 Simplify   ———————————
            t2 + 2t - 3

Trying to factor by splitting the middle term

 1.1     Factoring  t² + 2t - 3 

The first term is,  t²  its coefficient is  1 .
The middle term is,  +2t  its coefficient is  2 .
The last term, "the constant", is  -3 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -3 = -3 

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

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -1  and  3 
                     t² - 1t + 3t - 3

Step-4 : Add up the first 2 terms, pulling out like factors :
                    t • (t-1)
              Add up the last 2 terms, pulling out common factors :
                    3 • (t-1)
Step-5 : Add up the four terms of step 4 :
                    (t+3)  •  (t-1)
             Which is the desired factorization

Equation at the end of step  1  :

  ((t2)-1)   (9t+9)    (2-t)   
  ————————+(———————•———————————)
  ((t2)-4)  (4t+12) (t+3)•(t-1)

Step  2  :

             9t + 9
 Simplify   ———————
            4t + 12

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   9t + 9  =   9 • (t + 1) 

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   4t + 12  =   4 • (t + 3) 

Equation at the end of step  4  :

  ((t2)-1)  9•(t+1)    (2-t)   
  ————————+(———————•———————————)
  ((t2)-4)  4•(t+3) (t+3)•(t-1)

Step  5  :

Multiplying Exponential Expressions :

 5.1    Multiply  (t+3)  by  (t+3) 

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

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

Equation at the end of step  5  :

  ((t2)-1)  9•(t+1)•(2-t)
  ————————+——————————————
  ((t2)-4) 4•(t+3)2•(t-1)

Step  6  :

            t2 - 1
 Simplify   ——————
            t2 - 4

Trying to factor as a Difference of Squares :

 6.1      Factoring:  t² - 1 

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 - B²

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 1 is the square of 1
Check :  t²  is the square of  t¹ 

Factorization is :       (t + 1)  •  (t - 1) 

Trying to factor as a Difference of Squares :

 6.2      Factoring:  t² - 4 

Check : 4 is the square of 2
Check :  t²  is the square of  t¹ 

Factorization is :       (t + 2)  •  (t - 2) 

Polynomial Long Division :

 6.3    Polynomial Long Division
Dividing :  t + 1 
                              ("Dividend")
By         :    t + 2    ("Divisor")

Quotient :  1 
Remainder :  -1 

Equation at the end of step  6  :

  (t+1)•(t-1)  9•(t+1)•(2-t)
  ———————————+——————————————
  (t+2)•(t-2) 4•(t+3)2•(t-1)

Step  7  :

Calculating the Least Common Multiple :

 7.1    Find the Least Common Multiple

      The left denominator is :       (t+2) • (t-2) 

      The right denominator is :       4 • (t+3)²  • (t-1) 

      Least Common Multiple:
      4 • (t+2) • (t-2) • (t+3)²  • (t-1) 

Calculating Multipliers :

 7.2    Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = 4•(t+3)² •(t-1)

   Right_M = L.C.M / R_Deno = (t+2)•(t-2)

Making Equivalent Fractions :

 7.3      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)²   and  (y²+y)/(y+1)³  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

   L. Mult. • L. Num.      (t+1) • (t-1) • 4 • (t+3)2 • (t-1)
   ——————————————————  =   ——————————————————————————————————
         L.C.M             4 • (t+2) • (t-2) • (t+3)2 • (t-1)

   R. Mult. • R. Num.       9 • (t+1) • (2-t) • (t+2) • (t-2)
   ——————————————————  =   ——————————————————————————————————
         L.C.M             4 • (t+2) • (t-2) • (t+3)2 • (t-1)

Adding fractions that have a common denominator :

 7.4       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 (t+1) • (t-1) • 4 • (t+3)2 • (t-1) + 9 • (t+1) • (2-t) • (t+2) • (t-2)            4t5+11t4+17t3-2t2-48t-36       
 ——————————————————————————————————————————————————————————————————————  =  —————————————————————————————————————
                   4 • (t+2) • (t-2) • (t+3)2 • (t-1)                       4 • (t+2) • (t-2) • (t2+6t+9) • (t-1)

Trying to factor by pulling out :

 7.5      Factoring:  4t⁵ + 11t⁴ + 17t³ - 2t² - 48t - 36 

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

Group 1:  -48t - 36 
Group 2:  4t⁵ + 11t⁴ 
Group 3:  17t³ - 2t² 

Pull out from each group separately :

Group 1:   (4t + 3) • (-12)
Group 2:   (4t + 11) • (t⁴)
Group 3:   (17t - 2) • (t²)


Looking for common sub-expressions :

Group 1:   (4t + 3) • (-12)
Group 3:   (17t - 2) • (t²)
Group 2:   (4t + 11) • (t⁴)

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 :

 7.6    Find roots (zeroes) of :       F(t) = 4t⁵ + 11t⁴ + 17t³ - 2t² - 48t - 36
Polynomial Roots Calculator is a set of methods aimed at finding values of  t  for which   F(t)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  t  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  4  and the Trailing Constant is  -36.

 The factor(s) are:

of the Leading Coefficient :  1,2 ,4
 of the Trailing Constant :  1 ,2 ,3 ,4 ,6 ,9 ,12 ,18 ,36

 Let us test ....

Note - For tidiness, printing of 25 checks which found no root was suppressed

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
   4t⁵ + 11t⁴ + 17t³ - 2t² - 48t - 36 
can be divided with  t + 1 

Polynomial Long Division :

 7.7    Polynomial Long Division
Dividing :  4t⁵ + 11t⁴ + 17t³ - 2t² - 48t - 36 
                              ("Dividend")
By         :    t + 1    ("Divisor")

Quotient :  4t⁴+7t³+10t²-12t-36  Remainder:  0 

Polynomial Roots Calculator :

 7.8    Find roots (zeroes) of :       F(t) = 4t⁴+7t³+10t²-12t-36

     See theory in step 7.6
In this case, the Leading Coefficient is  4  and the Trailing Constant is  -36.

 The factor(s) are:

of the Leading Coefficient :  1,2 ,4
 of the Trailing Constant :  1 ,2 ,3 ,4 ,6 ,9 ,12 ,18 ,36

 Let us test ....

Note - For tidiness, printing of 25 checks which found no root was suppressed

Polynomial Roots Calculator found no rational roots

Trying to factor by splitting the middle term

 7.9     Factoring  t²+6t+9 

The first term is,  t²  its coefficient is  1 .
The middle term is,  +6t  its coefficient is  6 .
The last term, "the constant", is  +9 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 9 = 9 

Step-2 : Find two factors of  9  whose sum equals the coefficient of the middle term, which is   6 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  3  and  3 
                     t² + 3t + 3t + 9

Step-4 : Add up the first 2 terms, pulling out like factors :
                    t • (t+3)
              Add up the last 2 terms, pulling out common factors :
                    3 • (t+3)
Step-5 : Add up the four terms of step 4 :
                    (t+3)  •  (t+3)
             Which is the desired factorization

Multiplying Exponential Expressions :

 7.10    Multiply  (t+3)  by  (t+3) 

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

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

Final result :

  (4t4+7t3+10t2+12t+36)•(t+1)
  ———————————————————————————
  4•(t+2)•(t+2)•(t+3)2•(t+1)