Factorising quartics

1
Factorising quartics
One solution of the quartic equation
z4 2z³ 2z² 10z 25 0
is z -2 i.
Solve the equation.
2
Factorising quartics
Since z -2 i is a solution,
another solution is the complex conjugate z -2
- i .
Therefore two factors of the quartic expression
are (z 2 i) and (z 2 i).
So a quadratic factor is (z 2 i)(z 2 i).
Multiplying out gives
(z 2)² - (i)²
z² 4x 4 - (-1)
z² 4x 5
3
Factorising quartics
Now you need to factorise the quartic expression
z4 2z³ 2z² 10z 25
into two quadratic factors, where one factor is
z² 4z 5.
4
Factorising polynomials
This PowerPoint presentation demonstrates three
methods of factorising a quartic into two
quadratic factors when you know one quadratic
factor.
Click here to see factorising by inspection
Click here to see factorising using a table
Click here to see polynomial division
5
Factorising by inspection
Write the unknown quadratic as az² bz c.
z4 2z³ 2z² 10z 25 (z² 4z 5)(az² bz
c)
6
Factorising by inspection
Imagine multiplying out the brackets. The only
way of getting a term in z4 is by multiplying z2
by az2, giving az4.
z4 2z³ 2z² 10z 25 (z² 4z 5)(az² bz
c)
So a must be 1.
7
Factorising by inspection
Imagine multiplying out the brackets. The only
way of getting a term in z4 is by multiplying z2
by az2, giving az4.
z4 2z³ 2z² 10z 25 (z² 4z 5)(1z² bz
c)
So a must be 1.
8
Factorising by inspection
Now think about the constant term. You can only
get a constant term by multiplying 5 by c, giving
5c.
z4 2z³ 2z² 10z 25 (z² 4z 5)(z² bz
c)
So c must be 5.
9
Factorising by inspection
Now think about the constant term. You can only
get a constant term by multiplying 5 by c, giving
5c.
z4 2z³ 2z² 10z 25 (z² 4z 5)(z² bz
5)
So c must be 5.
10
Factorising by inspection
Now think about the term in z. When you multiply
out the brackets, you get two terms in z.
4z multiplied by 5 gives 20z
z4 2z³ 2z² 10z 25 (z² 4z 5)(z² bz
5)
5 multiplied by bz gives 5bz
So 20z 5bz 10z
therefore b must be -2.
11
Factorising by inspection
Now think about the term in z. When you multiply
out the brackets, you get two terms in z.
4z multiplied by 5 gives 20z
z4 2z³ 2z² 10z 25 (z² 4z 5)(z² - 2z
5)
5 multiplied by bz gives 5bz
So 20z 5bz 10z
therefore b must be -2.
12
Factorising by inspection
You can check by looking at the z² term. When you
multiply out the brackets, you get three terms in
z².
z² multiplied by 5 gives 5z²
z4 2z³ 2z² 10z 25 (z² 4z 5)(z² - 2z
5)
4z multiplied by -2z gives -8z²
5 multiplied by z² gives 5z²
5z² - 8z² 5z² 2z² as it should be!
13
Factorising by inspection
Now you can solve the equation by applying the
quadratic formula to z²- 2z 5 0.
z4 2z³ 2z² 10z 25 (z² 4z 5)(z² - 2z
5)
The solutions of the equation are z -2 i, -2
- i, 1 2i, 1 2i.
14
Factorising polynomials
Click here to see this example of factorising by
inspection again
Click here to see factorising using a table
Click here to see polynomial division
Click here to end the presentation
15
Factorising using a table
If you find factorising by inspection difficult,
you may find this method easier.
Some people like to multiply out brackets using a
table, like this
2x³
-6x²
-8x
3x²
-9x
-12
So (2x 3)(x² - 3x 4) 2x³ - 3x² - 17x - 12
The method you are going to see now is basically
the reverse of this process.
16
Factorising using a table
Write the unknown quadratic as az² bz c.
17
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
The only z4 term appears here,
so this must be z4.
18
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
This means that a must be 1.
19
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
This means that a must be 1.
20
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
25
The constant term, 25, must appear here
21
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
25
so c must be 5
22
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
25
so c must be 5
23
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
5z²
4z³
20z
25
5z²
Four more spaces in the table can now be filled in
24
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
-2z³
5z²
4z³
20z
25
5z²
This space must contain an z³ term
and to make a total of 2z³, this must be -2z³
25
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
-2z³
5z²
4z³
20z
25
5z²
This shows that b must be -2
26
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
-2z³
5z²
4z³
20z
25
5z²
This shows that b must be -2
27
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
-2z³
5z²
4z³
20z
-8z²
25
5z²
-10z
Now the last spaces in the table can be filled in
28
Factorising using a table
The result of multiplying out using this table
has to be z4 2z³ 2z² 10z 25
z4
-2z³
5z²
4z³
20z
-8z²
25
5z²
-10z
and you can see that the term in z²is 2z² and the
term in z is 10z, as they should be.
29
Factorising by inspection
Now you can solve the equation by applying the
quadratic formula to z²- 2z 5 0.
z4 2z³ 2z² 10z 25 (z² 4z 5)(z² - 2z
5)
The solutions of the equation are z -2 i, -2
- i, 1 2i, 1 2i.
30
Factorising polynomials
Click here to see this example of factorising
using a table again
Click here to see factorising by inspection
Click here to see polynomial division
Click here to end the presentation
31
Algebraic long division
Divide z4 2z³ 2z² 10z 25 by z² 4z 5
z² 4z 5 is the divisor
z4 2z³ 2z² 10z 25 is the dividend
The quotient will be here.
32
Algebraic long division
First divide the first term of the dividend, z4,
by z² (the first term of the divisor).
z²
This gives z². This will be the first term of the
quotient.
33
Algebraic long division
z²
z4 4z³ 5z²
Now multiply z² by z² 4z 5
-2z³ - 3z²
and subtract
34
Algebraic long division
z²
10z
z4 4z³ 5z²
-2z³ - 3z²
Bring down the next term, 10z
35
Algebraic long division
z²
- 2z
z4 4z³ 5z²
Now divide -2z³, the first term of -2z³ - 3z²
5, by z², the first term of the divisor
-2z³ - 3z²
10z
which gives -2z
36
Algebraic long division
z²
- 2z
z4 4z³ 5z²
-2z³ - 3z²
10z
Multiply -2z by z² 4z 5
-2z³- 8z²- 10z
5z² 20z
and subtract
37
Algebraic long division
z²
- 2z
25
z4 4z³ 5z²
-2z³ - 3z²
10z
Bring down the next term, 25
-2z³- 8z²- 10z
5z² 20z
38
Algebraic long division
z²
- 2z
5
z4 4z³ 5z²
Divide 5z², the first term of 5z² 20z 25, by
z², the first term of the divisor
-2z³ - 3z²
10z
-2z³- 8z²- 10z
5z² 20z
25
which gives 5
39
Algebraic long division
z²
- 2z
5
z4 4z³ 5z²
-2z³ - 3z²
10z
Multiply z² 4z 5 by 5
-2z³- 8z²- 10z
5z² 20z
25
Subtracting gives 0 as there is no remainder.
5z² 20z 25
0
40
Factorising by inspection
Now you can solve the equation by applying the
quadratic formula to z²- 2z 5 0.
z4 2z³ 2z² 10z 25 (z² 4z 5)(z² - 2z
5)
The solutions of the equation are z -2 i, -2
- i, 1 2i, 1 2i.
41
Factorising polynomials
Click here to see this example of polynomial
division again
Click here to see factorising by inspection
Click here to see factorising using a table
Click here to end the presentation