1. 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.

2. 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.

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

4. 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

5. 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

6. 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.

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

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

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

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

11. 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

12. 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

13. 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.

14. 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

15. 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

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

17. 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?

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

19. 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  

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

21. 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

22. 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

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

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

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

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

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

28. 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

29. The Cubic Rules

30. The Cubic Rules

31. The Cubic Rules

32. The Cubic Rules

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

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

35. 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.

36. 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
= –
= 0
so x – 1 is a factor of P(x).

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

38. Factorising Only Questions
Ex 7D p221
Q3 abde

39. Divisions With Remainders

40. Divisions With Remainders2 -5  2   

41. Remainders
Do Ex 7B p215
Q1 LHS
Q3a

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

43. Solving Cubic Equations Revision