Solving Cubic equations or polynomials of greater order

Quadratic

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.

Factorise

  X² - 6x + 5

 (x – 5) (x – 1) = 0

x = 5 or x = 1

Solve using formula

• b +/- √ b² - 4ac

2a

+ 6 +/- √ 36 – 4 x 1 x -5

                2 x 1

+ 6 + √ 56        or         + 6 - √ 56

          2                               2

Cubic

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

x + 1 √ x³ - 4x² + x + 6

        - x³ + x²

              - 5x² + x

            - - 5x² - 5x

                      6x + 6

                   -  6x + 6

(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

2. 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).

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.