Solution - Finding the roots of polynomials

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                (a+4)*(2*a-3)*(3*a-10)-(80)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((a + 4) • (2a - 3) • (3a - 10)) -  80  = 0 

Step  2  :

Equation at the end of step  2  :

  (a + 4) • (2a - 3) • (3a - 10) -  80  = 0 

Step  3  :

Checking for a perfect cube :

 3.1    6a³-5a²-86a+40  is not a perfect cube

Trying to factor by pulling out :

 3.2      Factoring:  6a³-5a²-86a+40 

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

Group 1:  -86a+40 
Group 2:  6a³-5a² 

Pull out from each group separately :

Group 1:   (43a-20) • (-2)
Group 2:   (6a-5) • (a²)

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(a) = 6a³-5a²-86a+40
Polynomial Roots Calculator is a set of methods aimed at finding values of  a  for which   F(a)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  a  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  6  and the Trailing Constant is  40.

 The factor(s) are:

of the Leading Coefficient :  1,2 ,3 ,6
 of the Trailing Constant :  1 ,2 ,4 ,5 ,8 ,10 ,20 ,40

 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
   6a³-5a²-86a+40 
can be divided with  a-4 

Polynomial Long Division :

 3.4    Polynomial Long Division
Dividing :  6a³-5a²-86a+40 
                              ("Dividend")
By         :    a-4    ("Divisor")

Quotient :  6a²+19a-10  Remainder:  0 

Trying to factor by splitting the middle term

 3.5     Factoring  6a²+19a-10 

The first term is,  6a²  its coefficient is  6 .
The middle term is,  +19a  its coefficient is  19 .
The last term, "the constant", is  -10 

Step-1 : Multiply the coefficient of the first term by the constant   6 • -10 = -60 

Step-2 : Find two factors of  -60  whose sum equals the coefficient of the middle term, which is   19 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  3  :

  (6a2 + 19a - 10) • (a - 4)  = 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.

Parabola, Finding the Vertex :

 4.2      Find the Vertex of   y = 6a²+19a-10

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 6 , is positive (greater than zero). 

 Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

 Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

 For any parabola,Aa²+Ba+C,the  a -coordinate of the vertex is given by  -B/(2A) . In our case the  a  coordinate is  -1.5833  

 Plugging into the parabola formula  -1.5833  for  a  we can calculate the  y -coordinate : 
  y = 6.0 * -1.58 * -1.58 + 19.0 * -1.58 - 10.0
or   y = -25.042

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 6a²+19a-10
Axis of Symmetry (dashed)  {a}={-1.58} 
Vertex at  {a,y} = {-1.58,-25.04} 
 a -Intercepts (Roots) :
Root 1 at  {a,y} = {-3.63, 0.00} 
Root 2 at  {a,y} = { 0.46, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.3     Solving   6a²+19a-10 = 0 by Completing The Square .

 Divide both sides of the equation by  6  to have 1 as the coefficient of the first term :
   a²+(19/6)a-(5/3) = 0

Add  5/3  to both side of the equation :
   a²+(19/6)a = 5/3

Now the clever bit: Take the coefficient of  a , which is  19/6 , divide by two, giving  19/12 , and finally square it giving  361/144 

Add  361/144  to both sides of the equation :
  On the right hand side we have :
   5/3  +  361/144   The common denominator of the two fractions is  144   Adding  (240/144)+(361/144)  gives  601/144 
  So adding to both sides we finally get :
   a²+(19/6)a+(361/144) = 601/144

Adding  361/144  has completed the left hand side into a perfect square :
   a²+(19/6)a+(361/144)  =
   (a+(19/12)) • (a+(19/12))  =
  (a+(19/12))²
Things which are equal to the same thing are also equal to one another. Since
   a²+(19/6)a+(361/144) = 601/144 and
   a²+(19/6)a+(361/144) = (a+(19/12))²
then, according to the law of transitivity,
   (a+(19/12))² = 601/144

We'll refer to this Equation as  Eq. #4.3.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (a+(19/12))²   is
   (a+(19/12))^2/2 =
  (a+(19/12))¹ =
   a+(19/12)

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:
   a+(19/12) = √ 601/144

Subtract  19/12  from both sides to obtain:
   a = -19/12 + √ 601/144

Since a square root has two values, one positive and the other negative
   a² + (19/6)a - (5/3) = 0
   has two solutions:
  a = -19/12 + √ 601/144
   or
  a = -19/12 - √ 601/144

Note that  √ 601/144 can be written as
  √ 601  / √ 144   which is √ 601  / 12

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    6a²+19a-10 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  a  , the solution for   Aa²+Ba+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  a =   ————————
                      2A

  In our case,  A   =     6
                      B   =    19
                      C   =  -10

Accordingly,  B²  -  4AC   =
                     361 - (-240) =
                     601

Applying the quadratic formula :

               -19 ± √ 601
   a  =    ——————
                      12

  √ 601   , rounded to 4 decimal digits, is  24.5153
 So now we are looking at:
           a  =  ( -19 ±  24.515 ) / 12

Two real solutions:

 a =(-19+√601)/12= 0.460

or:

 a =(-19-√601)/12=-3.626

Solving a Single Variable Equation :

 4.5      Solve  :    a-4 = 0 

 Add  4  to both sides of the equation : 
                      a = 4

Three solutions were found :

1.  a = 4
2.  a =(-19-√601)/12=-3.626
3.  a =(-19+√601)/12= 0.460