Expert: Scott A Wilson - 2/6/2009

Question
Hello, i need some help please...

Use the cubic formula to solve this:

x^3 + 4x^2 - 5x -3

Thanks

Answer
The cubic formula to solve ax3+bx2+cx+d=0 is

x = {q + [q² + (r-p²)^3]^0.5}^(1/3) +
{q - [q² + (r-p²)^3]^0.5}^(1/3) + p

where p = -b/(3a),  q = p^3 + (bc-3ad)/(6a²),  r = c/(3a).

We know that a=1, b=4, c=-5, and d=-3.
Using these, we can say that
p=-4/(3*1), q=-4^3/3^3 + (-20--9)/6, and r = -5/(3*1).

These can be reduced to
p = -4/3 and r = -5/3.

Now q may take a litte bit more.
We know that 4^3 = 64 and 3^3 = 27 and -20--9 = -11, so
q = -64/27 - 11/6.  The common denominator is 54.
q = -128/54 - 99/54 = -227/54.

Now that we have p, q, and r, we can apply the formula for x.

All that I have left to say is that
p = -4/3, q = -227/54, and r = -5/3 and that the formula for x is
x = {q + [q² + (r-p²)^3]^0.5}^(1/3) +
{q - [q² + (r-p²)^3]^0.5}^(1/3) + p,
which I could leave for you to grind through and come up with
an answer.  Note that (-227/54)² = 51,529/2,916.  You will get some
answer.  Call this answer E (why not?  That's a good letter...).

Once this is gotten, divide the cubic by (x-E)(quadratic).
Using the quadratic algorithm, you can find the roots of
what's left over.  Now that’s a long cumbersome way to do it.

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

Another way to do it would be in Excel and use the bisection method.

I can see that the roots are as follows:
1) between -5 and -4,
2) between -1 and 0, and
3) between 1 and 2.

The bisection method can be used on each of these sets.
To do so on the first set, start with xl = -5 (y=-3) and
xr -4 (y=17).  Evaluate the midpoint, -4.5.  

Since the value comes out to 9.375 and xr is also positve,
replace xr with -4.5.  Leave xl as -5, which has a functional
value of -3.

The average of -4.5 and -5 is -4.75.  
Evaluate the function at -4.75.  I get 3.828125, which is positive.  
Now since this value is positive at this point, replace -4.5 with
-4.75 since the value of the function at -4.75 is also positive.

The best way I've found to do this is put the first x value in A1.
In B1, put the functional value of A1, =((A1+4)*A1-5)*A1-3

In A2, put the second value for x.  Copy B1 to B2.
The functional value for A2 should be here.

In A3, put =(A1+A2)/2.  Copy B2 to B3.

In A3, do a control-C to copy the value.
Look at B3.  If B1 is the same sign, move to A1.
If B2 is the same sign, move to A2.
Once there, do an Edit - Paste Special - Value.

This will give you a new value in A3.
Repeat until as accurate as desired.

Here is my iteration for the first value:
The cubic formula to solve ax3+bx2+cx+d=0 is


Another way to do it would be in Excel and use the bisection method.

I can see that the roots are as follows:
1) between -5 and -4,
2) between -1 and 0, and
3) between 1 and 2.

The bisection method can be used on each of these sets.
To do so on the first set, start with xl = -5 (y=-3) and
xr -4 (y=17).  Evaluate the midpoint, -4.5.  

Since the value comes out to 9.375 and xr is also positve,
replace xr with -4.5.  Leave xl as -5, which has a functional
value of -3.

The average of -4.5 and -5 is -4.75.  
Evaluate the function at -4.75.  I get 3.828125, which is positive.  
Now since this value is positive at this point, replace -4.5 with
-4.75 since the value of the function at -4.75 is also positive.

The best way I've found to do this is put the first x value in A1.
In B1, put the functional value of A1, =((A1+4)*A1-5)*A1-3

In A2, put the second value for x.  Copy B1 to B2.
The functional value for A2 should be here.

In A3, put =(A1+A2)/2.  Copy B2 to B3.

In A3, do a control-C to copy the value.
Look at B3.  If B1 is the same sign, move to A1.
If B2 is the same sign, move to A2.
Once there, do an Edit - Paste Special - Value.

This will give you a new value in A3.
Repeat until as accurate as desired.

Here is my iteration for the first value:
-5         -3
-4         17
-4.5         9.375
-4.75         3.828125
-4.875         0.580078125
-4.9375         -1.167724609
-4.90625      -0.283355713
-4.890625      0.150966644
-4.8984375      -0.065541744
-4.89453125      0.042875469
-4.896484375      -0.011292361
-4.895507813      0.015801745
-4.895996094      0.002257241
-4.896240234      -0.004516923
-4.896118164      -0.001129682
-4.896057129      0.000563819
-4.896087646      -0.000282921
-4.896072388      0.000140451
-4.896080017      -7.12344E-05
-4.896076202      3.46086E-05
-4.89607811      -1.83129E-05
-4.896077156      8.14789E-06
-4.896077633      -5.08249E-06
-4.896077394      1.5327E-06
-4.896077514      -1.77489E-06
-4.896077454      -1.21095E-07
-4.896077424      7.05803E-07
-4.896077439      2.92354E-07
-4.896077447      8.56297E-08
-4.89607745      -1.77326E-08
-4.896077449      3.39485E-08
-4.896077449      8.10795E-09
-4.89607745      -4.81233E-09
-4.89607745      1.64781E-09

I conclude that the first value is -4.89607745.  These value are
hard to read since they are grouped together on the recieving end.

Knowing that would be a very messy number to try and factor,
I would repeat the process -1 and 0 as the first two x values.

Once this root has been found, the starting values should be
1 and 2.  Once this has been done, you now have the three roots.

Now that you have done all this, are you sure you wrote that
formula x^3 + 4x^2 - 5x -3 down correctly?