Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

PROGRAMME F6

POLYNOMIAL EQUATIONS

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Polynomial equations

Quadratic equations

Solution of cubic equations having at least one linear factor

Solution of fourth-order equations having at least two linear factors

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Polynomial equations

Quadratic equations

Solution of cubic equations having at least one linear factor

Solution of fourth-order equations having at least two linear factors

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Polynomial equations

In Programme F3 a polynomial in the variable x was evaluated by substituting the x-value into the equation and finding the resulting value for the polynomial expression.

This process is known as evaluating the expression.

Here the process is reversed by giving the polynomial expression the value of zero and finding those values of x which satisfy the resulting equation.

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Polynomial equations

Quadratic equations

Solution of cubic equations having at least one linear factor

Solution of fourth-order equations having at least two linear factors

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Polynomial equations

Quadratic equations

Solution of cubic equations having at least one linear factor

Solution of fourth-order equations having at least two linear factors

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Quadratic equations, ax2 + bx + c = 0

Solution by factors

Solution by completing the square

Solution by formula

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Quadratic equations, ax2 + bx + c = 0

Solution by factors

Where simple factors exist the solution can be derived from those. For example:

x2 + 5x – 14 can be factorized as (x + 7)(x – 2)so if:

x2 + 5x – 14 = 0 then (x + 7)(x – 2) = 0 and so x = −7 or x = 2

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Quadratic equations, ax2 + bx + c = 0

Solution by completing the square

Where simple factors do not exist the solution can be derived from completing the square. For example to solve x2 – 6x – 4 = 0 it is noted that x2 – 6x – 4 does not have simple factors so add 4 to both sides to give:

x2 – 6x = 4

Now, add the square of half the x-coefficient to both sides to give:

x2 – 6x + (–3)2 = 4 + (–3)2 that is x2 – 6x + 9 = (x – 3)2 = 13

Therefore x – 3 = ±√13 so x = 6.606 or x = −0.606 to 3 dp

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Quadratic equations, ax2 + bx + c = 0

Solution by formula

To solve ax2 + bx + c = 0 use can be made of the formula:

2 42

b b acx a

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Polynomial equations

Quadratic equations

Solution of cubic equations having at least one linear factor

Solution of fourth-order equations having at least two linear factors

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Polynomial equations

Quadratic equations

Solution of cubic equations having at least one linear factor

Solution of fourth-order equations having at least two linear factors

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Solution of cubic equations having at least one linear factor

In Programme F3 cubic polynomials were factorized with application of the remainder theorem and the factor theorem and the evaluation of polynomials by nesting.

These methods are reapplied to solve cubic equations.

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Polynomial equations

Quadratic equations

Solution of cubic equations having at least one linear factor

Solution of fourth-order equations having at least two linear factors

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Polynomial equations

Quadratic equations

Solution of cubic equations having at least one linear factor

Solution of fourth-order equations having at least two linear factors

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Solution of fourth-order equations having at least two linear factors

In Programme F3 fourth-order polynomials were factorized with application of the remainder theorem and the factor theorem and the evaluation of polynomials by nesting.

These methods are reapplied to solve fourth-order equations.

Programme F6: Polynomial equations

Worked examples and exercises are in the textSTROUD

Learning outcomes

Solve quadratic equations by factors, completing the square and by formula

Solve cubic equations with at least one linear factor

Solve fourth-order equations with at least two linear factors