Transcript Factorising polynomials - Further Mathematics Support

Factorising quartics
The example in this presentation is from
Example 2.10 in the FP1 textbook.
The aim is to factorise the quartic expression
z4 + 2z³ + 2z² + 10z + 25
into two quadratic factors, where one
factor is z² + 4z + 5.
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
Factorising by inspection
Write the unknown quadratic as
az² + bz + c.
z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(az² + bz + c)
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.
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.
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.
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.
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)
So 20z + 5bz = 10z
therefore b must be -2.
5 multiplied by bz
gives 5bz
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)
So 20z + 5bz = 10z
therefore b must be -2.
5 multiplied by bz
gives 5bz
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 5 multiplied by
-2z gives -8z²
z² gives 5z²
5z² - 8z² + 5z² = 2z²
as it should be!
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 + j, -2 - j, 1 + 2j, 1 – 2j.
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
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:
x²
2x 2x³
3 3x²
-3x
-6x²
-4
-8x
-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.
Factorising using a table
Write the unknown quadratic as az² + bz + c.
az²
z²
4z
5
bz
c
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
az²
z4
bz
c
4z
5
The only z4 term appears here,
so this must be z4.
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
az²
z4
bz
c
4z
5
This means that a must be 1.
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
1z²
z²
z4
bz
c
4z
5
This means that a must be 1.
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
z²
z4
bz
c
4z
5
25
The constant term, 25,
must appear here
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
z²
z4
bz
c
4z
5
25
so c must be 5
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
z²
z4
bz
5
4z
5
25
so c must be 5
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
z²
z4
4z 4z³
5
5z²
bz
5
5z²
20z
25
Four more spaces in the
table can now be filled in
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
z²
z4
4z 4z³
5
5z²
bz
-2z³
5
5z²
20z
25
This space must contain an z³ term
and to make a total of 2z³, this
must be -2z³
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
z²
z4
4z 4z³
5
5z²
bz
-2z³
5
5z²
20z
25
This shows that b must be -2
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
z²
z4
4z 4z³
5
5z²
-2z
-2z³
5
5z²
20z
25
This shows that b must be -2
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
z4
-2z
-2z³
5z²
4z 4z³
-8z²
20z
5
-10z
25
z²
5z²
5
Now the last spaces in the table can
be filled in
Factorising using a table
The result of multiplying out using this
table has to be z4 + 2z³ + 2z² + 10z + 25
z²
z4
-2z
-2z³
5z²
4z 4z³
-8z²
20z
5
-10z
25
z²
5z²
5
and you can see that the term in z²is 2z²
and the term in z is 10z, as they should be.
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 + j, -2 - j, 1 + 2j, 1 – 2j.
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
Algebraic long division
Divide z4 + 2z³+ 2z² + 10z + 25 by z² + 4z + 5
z
2
 4z  5 z
4
 2z
3
 2z
2
 10 z  2 5
z² + 4z + 5 is
the divisor
The quotient
will be here.
z4 + 2z³ + 2z² + 10z + 25
is the dividend
Algebraic long division
First divide the first term of the dividend, z4,
by z² (the first term of the divisor).
z²
z
2
 4z  5 z
This gives z².
This will be the
first term of
the quotient.
4
 2z
3
 2z
2
 10 z  2 5
Algebraic long division
z²
z
2
 4z  5 z
Now multiply z²
by z² + 4z + 5
and subtract
4
 2z
3
 2z
2
z4 + 4z³ + 5z²
-2z³ - 3z²
 10 z  2 5
Algebraic long division
z²
z
2
+ 10zz  2 5
z4 + 4z³ + 5z²
 4z  5 z
Bring down the
next term, 10z
4
 2z
3
 2z
2
-2z³ - 3z²
Algebraic long division
z ² - 2z
z
2
 4z  5 z
Now divide -2z³,
the first term of
-2z³ - 3z² + 5, by
z², the first term
of the divisor
4
 2z
3
 2z
2
z4 + 4z³ + 5z²
which gives -2z
 10 z  2 5
-2z³ - 3z²+ 10z
Algebraic long division
z ² - 2z
z
2
 4z  5 z
Multiply -2z by
z² + 4z + 5
and subtract
4
 2z
3
 2z
2
z4 + 4z³ + 5z²
 10 z  2 5
-2z³ - 3z²+ 10z
-2z³- 8z²- 10z
5z²+ 20z
Algebraic long division
z ² - 2z
z
2
 4z  5 z
4
 2z
3
 2z
2
z4 + 4z³ + 5z²
 10 z +
2
5
25
-2z³ - 3z²+ 10z
Bring down the
next term, 25
-2z³- 8z²- 10z
5z²+ 20z
Algebraic long division
z ² - 2z + 5
z
2
 4z  5 z
Divide 5z², the
first term of
5z² + 20z + 25, by
z², the first term
of the divisor
which gives 5
4
 2z
3
 2z
2
z4 + 4z³ + 5z²
 10 z  2 5
-2z³ - 3z²+ 10z
-2z³- 8z²- 10z
5z²+ 20z + 25
Algebraic long division
z ² - 2z + 5
z
2
 4z  5 z
4
 2z
3
 2z
2
z4 + 4z³ + 5z²
 10 z  2 5
-2z³ - 3z²+ 10z
Multiply z² + 4z + 5
-2z³- 8z²- 10z
by 5
5z²+ 20z + 25
Subtracting gives 0 as
5z² + 20z + 25
there is no remainder.
0
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 + j, -2 - j, 1 + 2j, 1 – 2j.
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