Question

2x cube -3x + X + 15 divided by 2x+3 is how much

Answer 1

 (2x3+5x2-28x-15)=0 Three solutions were found : x = 3 x = -5 x = -1/2 = -0.500

Step by step solution :Step  1  :Equation at the end of step  1  : (((2 • (x3)) + 5x2) - 28x) - 15 = 0 Step  2  :Equation at the end of step  2  : ((2x3 + 5x2) - 28x) - 15 = 0 Step  3  :Checking for a perfect cube :

 3.1    2x3+5x2-28x-15  is not a perfect cube 

Trying to factor by pulling out :

 3.2      Factoring:  2x3+5x2-28x-15 

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

Group 1:  -28x-15 
Group 2:  2x3+5x2 

Pull out from each group separately :

Group 1:   (28x+15) • (-1)
Group 2:   (2x+5) • (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 :

 3.3    Find roots (zeroes) of :       F(x) = 2x3+5x2-28x-15
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  2  and the Trailing Constant is  -15. 

 The factor(s) are: 

of the Leading Coefficient :  1,2 
 of the Trailing Constant :  1 ,3 ,5 ,15 

 Let us test ....

  P  Q  P/Q  F(P/Q)   Divisor     -1     1      -1.00      16.00        -1     2      -0.50      0.00    2x+1      -3     1      -3.00      60.00        -3     2      -1.50      31.50        -5     1      -5.00      0.00    x+5      -5     2      -2.50      55.00        -15     1     -15.00     -5220.00        -15     2      -7.50      -367.50        1     1      1.00      -36.00        1     2      0.50      -27.50        3     1      3.00      0.00    x-3      3     2      1.50      -39.00        5     1      5.00      220.00        5     2      2.50      -22.50        15     1      15.00      7440.00        15     2      7.50      900.00   

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 
   2x3+5x2-28x-15 
can be divided by 3 different polynomials,including by  x-3 

Polynomial Long Division :

 3.4    Polynomial Long Division 
Dividing :  2x3+5x2-28x-15 
                              ("Dividend")
By         :    x-3    ("Divisor")

dividend  2x3 + 5x2 - 28x - 15 - divisor * 2x2   2x3 - 6x2     remainder    11x2 - 28x - 15 - divisor * 11x1     11x2 - 33x   remainder      5x - 15 - divisor * 5x0       5x - 15 remainder       0

Quotient :  2x2+11x+5  Remainder:  0 

Trying to factor by splitting the middle term

 3.5     Factoring  2x2+11x+5 

The first term is,  2x2  its coefficient is  2 .
The middle term is,  +11x  its coefficient is  11 .
The last term, "the constant", is  +5 

Step-1 : Multiply the coefficient of the first term by the constant   2 • 5 = 10 

Step-2 : Find two factors of  10  whose sum equals the coefficient of the middle term, which is   11 .

     -10   +   -1   =   -11     -5   +   -2   =   -7     -2   +   -5   =   -7     -1   +   -10   =   -11     1   +   10   =   11   That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  1  and  10 
                     2x2 + 1x + 10x + 5

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

Equation at the end of step  3  : (2x + 1) • (x + 5) • (x - 3) = 0 Step  4  :Theory - Roots of a product :

 4.1    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 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.