Solving Cubics Starting Problem solve New Terminology – P(x) represents polynomial in x Using the P(x) terminology the problem above can be written as If P(x) = x3 – 2x2 – 5x + 6 find the values for x that make such that P(x) = 0 Before we start this topic I want to introduce the terminology P(x) = meaning polynomial in x This makes it easy to talk about cubics and other polynomials.

The old Trial and Error method. Solving Cubics Unlike quadratic equations there is no formula available for helping us to solve cubics and as yet we don’t have an algebraic method so we have to go back to The old Trial and Error method.

Solving Cubics Starting Problem Let’s try x = 1 So x = 1 is a solution because it makes the starting equation work out correctly

Solving Cubics Starting Problem Let’s try x = 2 Solutions so far x = 1 Let’s try x = 2 So x = 2 is not a solution because it doesn’t satisfy the starting equality

Solving Cubics Starting Problem Let’s try x = 3 Solutions so far x = 1 Let’s try x = 3 So x = 3 is a solution because it makes the starting equation work out correctly

Solving Cubics Starting Problem Should we try x = 4 x = 1 x = 3 Hint: Solutions so far x = 1 x = 3 Highest order term Constant Should we try x = 4 Hint: When using Trial & Error we usually try the whole number factors of the Constant first particularly when the highest order term has a coefficient of 1.

Solving Cubics Starting Problem Let’s try x = -1 Solutions so far x = 1 x = 3 Let’s try x = -1 So x = -1 is not a solution because it doesn’t satisfy the starting equality

Solving Cubics Starting Problem Let’s try x = -2 Solutions so far x = 1 x = 3 Let’s try x = -2 So x = -2 is a solution because it makes the starting equation work out correctly

Solving Cubics Starting Problem So now we have 3 solutions x = 1 x = 3 x = -2 Which is all we can expect because the maximum number of solutions for any polynomial = the order (maximum power) of the equation

There has got to be a quicker method!! Solving Cubics ARRRRGHHHHHH!!!!!!!!!!!!!!!! There has got to be a quicker method!!

Solving Cubics With Terminology Method 1: Use Main Application Write equation Highlight it Go to Interactive then Equation/Inequality then Solve Method 2: Use Graphs & Tables Application Enter polynomial as y1 Draw the graph and make sure all x intercepts are in the window Go to Analysis/G solve/Root

Solving Cubics With Terminology Method 1: Use Main Application Write equation Highlight it Go to Interactive then Equation/Inequality then Solve Method 2: Use Graphs & Tables Application Enter polynomial as y1 Draw the graph and make sure all x intercepts are in the window Go to Analysis/G solve/Root Use two technology methods to find the values for x that make such that P(x) = 0 if: P(x) = x3 – 2x2 – 5x + 6 x = -2, 1, 3

Solving Cubics With Algebra There is also an Algebraic method that can be used if technology is not available and it is sometimes quicker than guessing the answers like we did at the start. It is definitely quicker for situations where your solution involves one or two big numbers and/or fraction answers.

Solving Cubics With Algebra The basis of this method involves trying to factorize the polynomial. We are going to look at how we can go from To Which gives us x = 1, x = 3 & x = -2 from our work with quadratics

Solving Cubics With Algebra The steps to the factorisation (algebraic) method for solving cubics are: Find one factor Use that factor to find its co-factor Factorise the polynomial and deduce the solutions from the factorised polynomial

Step 1: Find A Factor Starting Problem Find A Factor Try x = 1 So x = 1 is a solution  (x - 1) is a factor

Step 2: Find Co-Factor If (x -1) is a factor then This is the co-factor and this is what we have to find What will be the order of this equation?

Step 2: Find Co-Factor One method to find a co-factor is to carry out the division Which should give

Step 2: Find Co-Factor 1 -1 -6 The RJ preferred method is to deduce the co-factor 1. Multiply co-factor blank by the factor (they have to multiply out to the polynomial) I call this the co-factor blank 2. Deduce co-factor co-efficients using the working space 1  -1  -6  

Step 3: Factorise Polynomial and deduce the solutions So x = 1 , -2 , 3

Example 2 Step 1: Find A Factor What are my best options for the next test Let’s try x = 1 So x = 1 does not lead to a factor

Example 2 Step 1: Find A Factor Let’s try x = -1 Because we already have the values but not the signs So x = -1 is a solution  (x + 1) is a factor

Example 2: Step 2: Find Co-Factor Setting out to find the co-factor 1 2  -8   

Example 2: Step 3 - Factorise Polynomial and deduce the solutions So x = -1 , -4 , 2

Examples To Try (x + 1)(x – 3)(x – 2) = 0  x = -1, x = 3, x = 2

Examples To Try (x – 1)(2x – 1)(x – 2) = 0  x = 1, x = 1/2, x = 2

Textbook Questions Exercise 7E p224 Q1 LHS (try RHS if you need extra practice) Q2 all parts

Some Tricks of The Trade For single x terms solve by undoing each process Do Ex 7E q4 bcde Read Example 16 Do Ex 7E q3 Read Example 17 Do Ex 7E q5 When asked to use technology use the graphic calculator as you did for the quadratics unit. Do Ex 7E q6

The Cubic Rules

The Cubic Rules

The Cubic Rules

The Cubic Rules

Questions using Cubic Rules Ex7D p221 Q4 all parts Answer Part A: Using the Cubic Formulae on the worksheet Cubic Equations - Formulae & Technology Exercises

Solving Cubics With Technology Answer Part B: Technology Techniques on the worksheet Cubic Equations - Formulae & Technology Exercises

The Factor Theorem We have been using a version of this for solving cubics The benefit of this new statement is that it extends our version to cover situations that are not equations such as straight factorisation problems.

Factor Proofs Example Without dividing, show that x – 1 is a factor of P(x) = 2x3 – 5x2 + x + 2. The factor theorem says that P(1) = 0 if x – 1 is a factor of P(x). P(1) = 2 x 13 – 5 x 12 + 1 + 2 = 2 – 5 + 1 + 2 = 0 so x – 1 is a factor of P(x).

Factor Proofs Ex 7Dp220 Q1 And for extension try Q2ab Q6

Factorising Only Questions Ex 7D p221 Q3 abde

Divisions With Remainders

Divisions With Remainders 2 -5  2   

Remainders Do Ex 7B p215 Q1 LHS Q3a

Remainder Theorem Do Ex 7C Q1 LHS Q2 From the Textbook See examples on the boards Do Ex 7C Q1 LHS Q2

Solving Cubic Equations Revision