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# The real number system

please how do I solve the equation in a real number system

4x^3-19x^2+19x+6=0

### 2 Answers by Expert Tutors

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
1
f(x) = 4x^3-19x^2+19x+6
Since f(3) = 0, x-3 is a factor. By synthetic division,
3|4 -19 19 6
......12 -21 6
------------------
...4 -7 -2 | 0
So, another factor is 4x^2-7x-2.
Solve 4x^2-7x-2 = (4x+1)(x-2) = 0
x = -1/4, 2
Answer: x = -1/4, 2, 3

How did you get the 3?
You can use rational root theorem, and by trial and error.
Nataliya D. | Patient and effective tutor for your most difficult subject.Patient and effective tutor for your mos...
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Let see if x = ±2, ±3, ±6 are root, because these are the simplest roots to test for.
4*23 - 19*22 + 19*2 + 6 = 0  ⇒  x1 = 2

Let's divide given polynomial by (x - 2)

4x2 - 11x - 3
(x - 2) | 4x3 - 19x2 + 19x + 6
-
4x3 - 8x2
- 11x2  + 19x
-
- 11x2 + 22x
- 3x + 6
-
- 3x + 6
0

(x - 2)(4x2 - 11x - 3) = 0

4x2 - 11x - 3 = 0
D = b2 - 4ac = (-11)2 - 4*4*(-3) = 169

- b ± √D
x23 = -------------
2a

-(-11) ± √169
x23 = -------------------
2*4

x2 = (11 + 13) / 8 = 3
x3 = (11 - 13) / 8 = - 2/8 = - 1/4

The roots of a given equation are {- 1/4, 2, 3} .