Help Resolve this math equation

Question

3x^3 + 4x^2 - 15x - 20

please show how you got the answer so I can get a clear understanding for future problems like this

Answer 1

Rational Root Theorem:

 

 P = {1,2,4,5,10,20} <--- factors of 20

 Q = {1,3}  <--- factors of 3

 

P/Q = {1, 2, 4, 5, 10, 20, 1/3,  2/3,  4/3,   5/3,  10/3, 20/3 }

 

Keep in mind that these sets ALSO INCLUDE NEGATIVE... -1, -2, -4, -5, etc

You have to plug each one of the candidates in P/Q into the polynomial

to check, verify, and test if any of them result in zero.

 

It turns out that -4/3 works.

3(-4/3)^3 + 4(-4/3)^2 - 15(-4/3) - 20 =

3(-64/27) + 4(16/9) + 15(4/3) - 20 =

(-64/9) + 64/9 + 60/3 - 20 = 0  <---- 64/9 cancels;  60/3 = 20 cancels

 

Synthetic division...

 

-4/3  |   3      4        -15      -20

                   -4          0       20

        ------------------------------ 

             3     0         -15      0

 

The solution x=-4/3 means that the factor is (3x + 4)

So the quadratic that results is 3x^2 + 15 after (3x+4) is factored;

 

3x^2 - 15 = 0

3(x^2 - 5) = 0

 

  x^2-5 = 0

 

   x^2 = 5

 

    x = +or- sqrt(5)

 

CHECK:

 

   x = sqrt(5) :  

           3(sqrt(5))^3 + 4*(sqrt(5))^2 - 15*sqrt(5)  - 20 =

         3* 5 * sqrt(5) + 4*5 - 15*sqrt(5) - 20 =

        15*sqrt(5) + 20 - 15*sqrt(5)  - 20 = 0

 

 

x = -sqrt(5) : (watch the signs!!!!)

 
3(-sqrt(5))^3 + 4*(-sqrt(5))^2 - 15*-sqrt(5) - 20 =
3* -5 * sqrt(5) + 4*5 + 15*sqrt(5) - 20 =    <--- italic term is the cube of a negative which is negative;

                                                                           bold term is negative times negative which is positive;
-15*sqrt(5) + 20 +15*sqrt(5) - 20 = 0

 

The three solutions are -4/3,  +or- sqrt(5)

Tried, tested, checked, verified, and proven to be correct.

   x = sqrt(5) :  

           3(sqrt(5))^3 + 4*(sqrt(5))^2 - 15*sqrt(5)  - 20 =

         3* 5 * sqrt(5) + 4*5 - 15*sqrt(5) - 20 =

        15*sqrt(5) + 20 - 15*sqrt(5)  - 20 = 0