Post on 26-Mar-2015
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Solving Cubics
Starting Problem
solve 0652 23 xxx
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.
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
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
0652 23 xxx
Let’s try x = 1
0
6521
615121)1( 23
P
So x = 1 is a solution because it makes the starting equation work out correctly
Solving Cubics
Starting Problem
0652 23 xxx
Let’s try x = 2
So x = 2 is not a solution because it doesn’t satisfy the starting equality
4
61088
625222)2( 23
P
Solutions so far
x = 1
Solving Cubics
Starting Problem
0652 23 xxx
Let’s try x = 3
So x = 3 is a solution because it makes the starting equation work out correctly
0
6151827
635323)3( 23
P
Solutions so far
x = 1
Solving Cubics
Starting Problem
0652 23 xxx
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.
Solutions so far
x = 1 x = 3
Highest order term Constant
Solving Cubics
Starting Problem
0652 23 xxx
Let’s try x = -1
So x = -1 is not a solution because it doesn’t satisfy the starting equality
8
6521
615)1(2)1()1( 23
P
Solutions so far
x = 1 x = 3
Solving Cubics
Starting Problem
0652 23 xxx
Let’s try x = -2
So x = -2 is a solution because it makes the starting equation work out correctly
0
61088
625)2(2)2()2( 23
P
Solutions so far
x = 1 x = 3
Solving Cubics
Starting Problem
0652 23 xxx
So now we have 3 solutions x = 1 x = 3 x = -2Which is all we can expect because the maximum number of solutions for any polynomial = the order (maximum power) of the equation
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.
0652 23 xxx
0)2)(3)(1( xxx
We are going to look at how we can go from
Which gives us x = 1, x = 3 & x = -2 from our work with quadratics
To
Solving Cubics With Algebra
The steps to the factorisation (algebraic) method for solving cubics are:
1. Find one factor
2. Use that factor to find its co-factor
3. Factorise the polynomial and deduce the
solutions from the factorised polynomial
Step 1: Find A Factor
Starting Problem
652)(
065223
23
xxxxPlet
xxxSolve
Find A Factor Try x = 1
0
6521
615121)1( 23
P
So x = 1 is a solution (x - 1) is a factor
Step 2: Find Co-Factor
If (x -1) is a factor then
__)__)(__1(
6522
23
xxx
xxx
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
6521
___
23
2
xxxx
xx Which should give
1
Step 2: Find Co-FactorThe RJ preferred method is to deduce the co-factor
)1(__66
)1(__1
)1(__1
6521
___
2
223
23
2
xx
xxxx
xxxx
xxxx
xx
-1
-6
62 xx
1. Multiply co-factor blank by the factor (they have to multiply out to the polynomial)
2. Deduce co-factor co-efficients using the working space
I call this the co-factor blank
0)3)(2)(1(
0)6)(1(
06522
23
xxx
xxx
xxx
Step 3: Factorise Polynomial and deduce the solutions
So x = 1 , -2 , 3
Example 2 Step 1: Find A Factor
Example 2
Let’s try x = 1
So x = 1 does not lead to a factor
0863 23 xxx
10
8631
816131)1( 23
P
What are my best options for the next
test
Example 2 Step 1: Find A Factor
Example 2
Let’s try x = -1
0863 23 xxx
So x = -1 is a solution (x + 1) is a factor
0
8631
8)1(6)1(3)1()1( 23
P
Because we already have
the values but not the signs
1
)1(__88
)1(__22
)1(__1
8631
2
223
23
xx
xxxx
xxxx
xxxx
Example 2: Step 2: Find Co-Factor
Setting out to find the co-factor
2
-8
822 xx
0)2)(4)(1(
0)82)(1(
08632
23
xxx
xxx
xxx
Example 2: Step 3 - Factorise Polynomial and deduce the solutions
So x = -1 , -4 , 2
Examples To Try
064:3 23 xxxEx
030114:4 23 xxxEx
0252:5 23 xxxEx
02021:6 3 xxEx
02:7 23 xxxEx
08168:8 23 xxxEx
(x + 1)(x – 3)(x – 2) = 0 x = -1, x = 3, x = 2
(x – 2)(x – 5)(x + 3) = 0 x = 2, x = 5, x = -3
(x – 1)(2x + 1)(x – 2) = 0 x = 1, x = -1/2, x = 2
(x – 1)(x + 5)(x – 4) = 0 x = 1, x = -5, x = 4
(x + 2)(x²- x + 1) = 0 x = -2
Examples To Try
0252:5 23 xxxEx
02021:6 3 xxEx
02:7 23 xxxEx
(x – 1)(2x – 1)(x – 2) = 0 x = 1, x = 1/2, x = 2
(x – 1)(x + 5)(x – 4) = 0 x = 1, x = -5, x = 4
(x + 2)(x²- x + 1) = 0 x = -2
53,53,2
0)53)(53)(2(
x
xxx
08168:8 23 xxxEx
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
))((
))((
33)(
33)(
2233
2233
32233
32233
babababa
babababa
babbaaba
babbaaba
The Cubic Rules
2754368
279233438
33)2(33)2(3)2(
3233)32(
23
23
3223
32233
xxx
xxx
xxx
bxababbaax
6414410827
64163349327
44)3(34)3(3)3(
4333)43(
23
23
3223
32233
xxx
xxx
xxx
bxababbaax
The Cubic Rules
)964)(32(
)33.2)2)((32(
32))((
3)2(278
2
22
22
333
xxx
xxx
bxabababa
xx
)252016)(54(
)55.4)4)((54(
54))((
5)4(12564
2
22
22
333
xxx
xxx
bxabababa
xx
The Cubic Rules
)93)(3(2
3))((2
)3(2
)27(2542
2
22
33
33
xxx
bxabababa
x
xx
Questions using Cubic Rules
Ex7D p221
Q4 all parts
Answer
Part A: Using the Cubic Formulae
on the worksheetCubic Equations - Formulae & Technology Exercises
Solving Cubics With Technology
Answer
Part B: Technology Techniques
on the worksheetCubic Equations - Formulae & Technology Exercises
The Factor Theorem
)()()(
0)(0
xPoffactoraisbaxsoandxPof
solutionaisa
bthen
a
bPfindweandxPIf
We have been using a version of this for solving cubics
0)()( a
bPthenxPoffactoraisbaxIf
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
1
4332 23
x
xxx
6
)1(__22
)1(__55
)1(__22
43321
2
223
23
xx
xxxx
xxxx
xxxx
Divisions With Remainders
-5
2
1
6252 2
xxx
2
Remainders
• Do Ex 7B p215
• Q1 LHS
• Q3a
Remainder Theorem
See examples on the boards
Do Ex 7C Q1 LHS Q2
From the Textbook
)()(b
aPisremainderthebaxbydividedisxPWhen
)()(a
bPisremainderthebaxbydividedisxPWhen
Solving Cubic Equations Revision