completing squares, but with cubes...?

Question

I know how to complete the square of a formula to find its roots, or zeroes. I am having difficulty doing the same for cube formulas. The specific question I'm stuck on is:
x^3 - 15x +1 =0 has 3 answers, but I have no idea about how to find the answers.
I know I could just graph it on my calculator, but I need to show my work. Any help would be much appreciated.

i think i'm supposed to use the intermediate value theorem, but I don't know how to use that theorem for 3 answers. Do I just keep trying out different pairs of numbers to sub in for x until I get 3 sets whose answers have zero in between them?
If so are there any pointers on how to pick sets of numbers.
I'm given the interval: [-4, 4], so I am assuming I should start with f(-4) and f(4)

Answer 1

F(x) = x^3 - 15x +1
given [-4, 4]

x F(x) Sign
-4 -64 + 60 + 1 = -3 -
-3 -27 + 45 + 1 = 19 +
-2 -8 + 30 + 1 = 25 +
-1 -1 + 15 + 1 = 15 +
0 0 - 0 + 1 = 1 +
1 1 - 15 + 1 = -13 -
2 8 - 30 + 1 = -21 -
3 27 - 45 + 1 = -14 -
4 64 - 60 + 1 = 5 +

because you will hit a F(x) = 0 everytime the sign changes,
there is a root between -4 and -3, a root between 0 and 1, and a root between 3 and 4.
you can use this method by changing the interval from 1 to .1 or .01 or smaller to exact your solutions

Answer 2

There is special method to solve this type of problems. From a general cubic equaion ax^3+bx^2+cx+d = 0, first we eleminate x^2 term by making the substitution x = y+r. Then the cubic equation becomes
y^3 +(3 r + b) y^2 +(3 r^2 + 2 r b + c) y + r^3 + r^2 b + r c + d = 0

Now we choose y such that the quadratic term disappears

choose r = -b/3
So, with the substitution x = y - b/3, then original cubic equation reduces to y^3 + e y + f = 0
Vieta's substitution:
We try Vieta-substitution to reduce the last equation.
y = z + s 1/z. The constant 's' is an undefined constant for the present. Then
(z + s/z)^3 + e (z + (s/z) + f = 0
Expanding and multiplying through by z^3 , we have
z^6 + (3 s + e) z^4 + f z^3 + s (3 s + e) z^2 + s^3 = 0
Now we choose s = -e/3.
The equation becomes z^6 + f z^3 - e^3/27 = 0.
With z^3 = u
u^2 + f u - e^3/27 = 0
This is an easy to solve quadratic equation. Once we solve for 'u', w can find z^3 that is 'z'.

But when I tried this method for the eq given by you we are getting complex roots. There is something wrong in copying down Please check up.
For further details, follow the following link or Higher Algebra by Hall & Knight

Answer 3

The only possible rational zeros are 1 and -1.
Plugging them in for x, you will see that neither are zeros.
x^3 - 15x +1 changes sign 2 times. That means there will be 2 or 0 positive zeros.
(-x)^3 - 15(-x) + 1 = -x^3 + 15x + 1 changes signs one time. That means there will be exactly one negative zero.
Cases:
1) 2 positive, 1 negative and no complex
2) 0 positive, 1 negative and 2 complex
I can't think of an easy way of finding them without using a graphing calculator.

Answer 4

I'm so sorry but you can't complete the square with a cube. That's why you should have listened when they explained all that stuff about factoring.

Ok here's what you do. Suppose x = 0. Well that doesn't work so we know that x is not 0. Now rearrange the equation.
x^3 - 15x = -1
Divide by x (we can do this since we know x <> 0)

x^2 - 15 = -1 / x

x^2 + 0 * x - 15 = -1/x

(x + sqr(15)) * (x - sqr(15)) = -1/x

x * (x + sqr(15)) * (x - sqr(15)) = -1

x * (x + sqr(15)) * (x - sqr(15)) + 1 = 0

(x^2 + sqr(15)x) * (x - sqr(15)) + 1 = 0

x^3 + sqr(15)*(x-1) + 15 + 1 = 0

original equation:
x^3 - 15x + 1 = 0
so

-15x = sqr(15)*(x-1) + 15

-15x - sqr(15)x = 15 - sqr(15)

x(-15 - sqr(15)) = 15 - sqr(15)

so,

x = [15 - sqr(15)] / [-15 - sqr(15)]

ok you do the math.

Answer 5

There is a famous recipe to solve equations of the form

[1] ... x^3 - 3px - q = 0

which in your case has p = 5 and q = -1.
It involves writing x = u + v and solving the system of equations

[2] ... u^3 + v^3 = q
[3] ... u^3 . v^3 = p^3

However, if the equation has three real solutions, this method results in complex values for u^3 and v^3, which is extremely unpractical. (If I am not mistaken, raobn ran into this problem when trying to answer your question.)

The alternative method generates the three solutions as "points on a circle". I will not prove it here, but state the result.

The radius of the circle is

[4] ... r = 2 sqrt(p)

-- in your case, r = 2V5 = 4.472. Now calculate an angle as follows

[5] ... C = q/[2p sqrt(p)]
[6] ... t = (inv cos C) / 3

You find C = -1/(10V5) = -0.04472. This is the cosine of 92.56 degrees, so we find t = 30.85 degrees. The solutions are now

[7] ... x1 = r cos t
[8] ... x2 = r cos (t + 120)
[9] ... x3 = r cos (t - 120)

With your numbers,

x1 = 4.472 cos 30.85 = 3.8392
x2 = 4.472 cos 150.85 = -3.9059
x3 = 4.472 cos (-89.15) = 0.0667

If you want more details on this and other problems involving cubic equations, contact me; I wrote an 18 page paper on methods to solve them and could e-mail them as a Postscript or PDF file.

Answer 6

Key: ^ means squared * means cubed 2x+3x^+x* factor out x > x(2+3x+x^) find common factors wich r 2 nd 1 2+2x+1x+x^ divide down da middle nd factor 2(1+x) x(1+x) (2+x) nd (1+x) now bring bac the fisrt x dat was factored x(2+x)(1+x) 2 check it multiply it through nd u get 2x+3x^+x* dats the anwer nd its complete cuz u cant do anythin more 2 it :)

Answer 7

x^3 - 15x= -1
(x+1) (x+5) (x-3) = -1
x=1,5, -3
plug each number in to "x" for 3 values.

Answer 8

x³ - 15x + 1 = (x - a) (x - b) (x - c) = x³ - (a+b+c)x² + (ab+ac+bc)x - abc

In your case:
a+b+c = 0
ab + ac + bc = 15
abc = -1

That will give you very interesting numbers, quite unlikely to be rational.

Answer 9

I'll go answering sports questions.

Answer 10

x = 0.0666864373322236
x = 3.8392095168440132
x = -3.905895954176237