• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Solving Cubic equations or polynomials of greater order

Extracts from this document...


Solving Cubic equations or polynomials of greater order


I firstly solved our quadratic equations using factorising, this worked well until I found equations that have factors that are not whole. I used the formula for these equations, although the formula is very hard to remember and if it’s written incorrectly the answer will be wrong, so I tried using graphs this was ok although the graphs are not very accurate, they are only around 1 decimal place.


  X² - 6x + 5

 (x – 5) (x – 1) = 0

x = 5 or x = 1

Solve using formula

  • b +/- √ b² - 4acimage00.png


+ 6 +/- √ 36 – 4 x 1 x -5image00.png

                2 x 1

+ 6 + √ 56        or         + 6 - √ 56image00.pngimage00.png

          2                               2


Here I first tried solving the equations by factorising, as with the quadratic equations this worked well until I came across equations with factors including decimal places. For these I tried as before to use the formula but I firstly need to find one of the factors, and if all factors include decimal places this can be difficult. So I lastly

...read more.


x + 1 √ x³ - 4x² + x + 6

        - x³ + x²

              - 5x² + ximage09.png

            - - 5x² - 5x

                      6x + 6image09.png

                   -  6x + 6image09.png

(x + 1) (x² - 5x + 6)

(x + 1) (x – 3) (x – 2)

Change of Sign

There are three types:-

  1. Interval Estimation which includes
  1. Decimal Search
  2. Interval Bisection
  1. Linear Interpolation

I have decided to use decimal search.

In order to use this method I must find an approximate root. I have already drawn the graph and so I can see that there is a root that lies between 1 and 2.

From my graph I know the root is 2.8 correct to one decimal place. With decimal search to find a more accurate value for the root I must look for a change of sign to locate the root. With decimal search after each iteration I divide the interval into 10 equal parts.

  • Step 1 – Once a root is found, you use increments of firstly 0.1 working out the value for the function. You do this until a change of sign is found.
  • Step 2 – There is a change of sign, therefore there is a root. Having narrowed down the interval, you continue with increments of 0.001. looking for a change of sign.
  • Step 3 – This process is continued to find a more accurate root, which is usually a value correct to 5 decimal places.

Whilst using the change of sign method to locate accurate values for roots I came across some equations where there is a root but the change of sign method does not show the root, this is because there is no change of sign. I feel this method is not reliable enough as it does not find all roots. I am now going to research other methods of finding roots.

Fixed point iteration

In fixed point iteration you find a single value as your estimate for the x value. This involves an interactive process a method of generating a sequence of numbers by continued repetition of the same procedure. If the numbers obtained in this manner approach some limiting value, they are said to converge to this value.

  • Step 1 – With the chosen equation, you must rearrange it in the form x = g ’ (x). This provides the basis for the iterative formula.
  • Step 2 – Choose a start value for x.
  • Step 3 – Find the corresponding value of g ‘ (x).
...read more.


x values oscillate about the root.

If the equation is arranged in some ways the equation will not converge as x = g ‘ (x)will only converge to a root if    -1 < g ‘ (x) < 1     for values of x close to the root.



Newton Raphson

Another fixed point estimation method, and as with the previous method it is necessary to use an estimate of the root as a starting point.

  • Step 1 - You start with an estimate, x1, for a root of f ’ (x) = 0.            
  • Step 2 – You then draw a tangent to the curve Y = f ‘ (x) at the point (x1, f ‘ (x)).
  • Step 3 – The point at which the tangent cuts the x axis then gives the next approximation for the root.
  • Step 4 – This process is then repeated until the values converge.


I have come across a few problems with this method also as if a poor starting value is chosen, the iteration may diverge. Or the tangent may meet the axis at a point outside the domain of the function. If this happens you will not locate the root, even if it is there.

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    C3 Coursework - different methods of solving equations.

    5 star(s)

    the point on the X-axis, and can do so to many places. This makes it much easier than excel but and is almost accurate as it is because it gives the root to 8+ decimal places and that is accurate.

  2. Solutions of equations

    So taking the answer as the midpoint of the two I get: x = 0.666665 (6 d.p.). But seeing as I only need the answer to 5 decimal places I get x = 0.66667 (5 d.p.). Answer: 0.66667 (5 d.p.)

  1. Mathematical equations can be solved in many ways; however some equations cannot be solved ...

    So we go across to the line y=x, which reflects this value onto the x-axis, and then down to the curve. This produces the value -3.382999. I then repeated the process, going across to the line and down to the curve each time.

  2. Numerical solutions of equations

    By looking at the diagram (see Figure 3), I can see that my starting value (x1) is -1.5. x1 = -1.5 x2 = -1.295 (I substituted x1 = -1.5 into the iterative formula to get the value for x2) x3 = -1.227690 (I substituted x2 = -1.295 into the iterative formula to get x3)

  1. Numerical Method (Maths Investigation)

    However, with or without the use of computer, Newton-Raphson Method seems to be the best method among the three method I have been used to compare here. It has an iteration formula: . I can find the gradient of the tangent to the curve at certain point by differentiate the equation of the curve and substitute the Xn into it.

  2. Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson ...

    -0.32 1.753E-06 6.10352E-05 15 -0.320251 1.75303E-06 -0.32019 -6.13511E-07 -0.32 5.6981E-07 3.05176E-05 16 -0.320221 5.69814E-07 -0.32019 -6.13511E-07 -0.32 -2.183E-08 1.52588E-05 17 -0.320221 5.69814E-07 -0.32021 -2.18343E-08 -0.32 2.7399E-07 7.62939E-06 18 -0.320213 2.73993E-07 -0.32021 -2.18343E-08 -0.32 1.2608E-07 3.8147E-06 19 -0.32021 1.2608E-07 -0.32021 -2.18343E-08 -0.32 5.2123E-08 1.90735E-06 20 -0.32021 5.21233E-08 -0.32021 -2.18343E-08 -0.32 1.5145E-08 9.53674E-07 Spreadsheet 1.8 Graph1.9--- bisection of y = (x+0.3)

  1. Numerical solution of equations

    Here are three steps of this method: - Take increment in x of size 0.1 within the interval of the root. Stop when there is a sign change. - The interval has been narrowed down and now keeps on taking increment in x of size 0.01 within the narrowed interval.

  2. C3 COURSEWORK - comparing methods of solving functions

    Therefore, the root is 0.87939 to 5 decimal places. Lastly, we will use x=g(x) method to find the root in the interval [0, 1]: I need to rearrange the equation x³+3x²–3=0 in to the form of x=g(x), Let X1= 1 x³+3x²–3=0 n x 1 1 0.816496581 2 0.816496581 0.904741021 3

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work