12x^3-65x^2+24x+10=0

Question

How do I find the roots if one of the roots is fraction 2/3

Answer 1

f(x) = 12x^3 – 65x^2 + 24x + 10 = 0

 

By the Factor Theorem of Polynomials we know:

 

f(x) = 12(x – r1)(x – r2)(x – r3)

 

Since one of the roots is 2/3:

 

f(x) = 12(x – 2/3)(x – r2)(x – r3) = 0

 

f(x)/(x – 2/3) = 12(x – r2)(x – r3) = 0

 

Using Synthetic Division:

 

2/3 | 12 –65  24    10

                8 –38 –28/3        

        12 –57 –14 | ≠0 SO NOT A FACTOR

 

2/3 is NOT a root of this equation.

 

g(x) = 12 (x + 1 / 4) (x - 2 / 3) (x - 5) probably what was intended (from zeros of GeoGebra graph of f(x): http://www.wyzant.com/resources/files/266857/student_gave_wrong_equation.)

 

g(x) = (4x + 1) (3x - 2) (x - 5)

= (12 x^2 – 5x – 2)(x - 5)

= 12 x^3 – 5x^2 – 2x – 60x^2 + 25x + 10

= 12x^3 – 65x^2 + 23x + 10

 

So coefficient of x should have been 23.

 

[Tutors should always check problem for consistency before investing a lot of time into an answer.]

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Let’s start fresh and assume it was g(x) that was given with root 2/3.

Since 2/3 is a root, x - 2/3 is a factor, and we can use Synthetic Division to find the product of the other two factors.

2/3 | 12 –65  23   10
                8 –38 –10
        12 –57 –15 | 0

g(x) = (x – 2/3)(12x^2 – 57x – 15)
g(x) = 3(x – 2/3)(4x^2 – 19x – 5)

4(-5) = -20

-1(20) 19
1(-20) -19 <==

4x^2 – 20x + 1x – 5 =
4x(x – 5) + 1(x – 5) =
(x – 5)(4x + 1)

g(x) = (3x – 2)(x – 5)(4x + 1)
which has zeros/roots of 2/3, 5, –1/4