factorising quartics
DESCRIPTION
Factorising quartics. One solution of the quartic equation. z 4 + 2 z ³ + 2 z ² + 10 z + 25 = 0. is z = -2 + i. Solve the equation. Factorising quartics. Since z = -2 + i is a solution,. another solution is the complex conjugate z = -2 - i. - PowerPoint PPT PresentationTRANSCRIPT
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Factorising quartics
One solution of the quartic equation
z4 + 2z³ + 2z² + 10z + 25 = 0
Solve the equation.
is z = -2 + i.
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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
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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.
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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
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Write the unknown quadratic as
az² + bz + c.
z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(az² + bz + c)
Factorising by inspection
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(az² + bz + c)
Imagine multiplying out the brackets. The only way of getting a term in z4 is by multiplying z2 by az2, giving az4.
So a must be 1.
Factorising by inspection
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(1z² + bz + c)
Imagine multiplying out the brackets. The only way of getting a term in z4 is by multiplying z2 by az2, giving az4.
So a must be 1.
Factorising by inspection
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² + bz + c)
Now think about the constant term. You can only get a constant term by multiplying 5 by c, giving 5c.
So c must be 5.
Factorising by inspection
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² + bz + 5)
Now think about the constant term. You can only get a constant term by multiplying 5 by c, giving 5c.
So c must be 5.
Factorising by inspection
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² + bz + 5)
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
5 multiplied by bz gives 5bz
So 20z + 5bz = 10z
therefore b must be -2.
Factorising by inspection
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)
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
5 multiplied by bz gives 5bz
So 20z + 5bz = 10z
therefore b must be -2.
Factorising by inspection
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)
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
4z multiplied by -2z gives -8z²5z² - 8z² + 5z² = 2z²
as it should be!
Factorising by inspection
5 multiplied by z² gives 5z²
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)
Factorising by inspection
Now you can solve the equation by applying the quadratic formula to z²- 2z + 5 = 0.
The solutions of the equation are z = -2 + i, -2 - i, 1 + 2i, 1 – 2i.
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Factorising polynomials
Click here to see this example of factorising by inspection again
Click here to see factorising using a table
Click here to end the presentation
Click here to see polynomial division
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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:
2x
3
x² -3x - 4
2x³ -6x² -8x
3x² -9x -12
So (2x + 3)(x² - 3x – 4) = 2x³ - 3x² - 17x - 12The method you are going to see now is basically the reverse of this process.
Factorising using a table
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Write the unknown quadratic as az² + bz + c.
Factorising using a table
z²
4z
5
az² bz c
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z²
4z
5
az² bz c
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
The only z4 term appears here,
so this must be z4.
z4
Factorising using a table
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z²
4z
5
az² bz c
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
This means that a must be 1.
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z²
4z
5
1z² bz c
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
This means that a must be 1.
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z²
4z
5
z² bz c
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
The constant term, 25, must appear here
25
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z²
4z
5
z² bz c
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
25
so c must be 5
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z²
4z
5
z² bz 5
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
25
so c must be 5
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z²
4z
5
z² bz 5
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
25
Four more spaces in the table can now be filled in
4z³
5z²
5z²
20z
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z²
4z
5
z² bz 5
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
25
4z³
5z²
5z²
20z
This space must contain an z³ termand to make a total of 2z³, this must be -2z³
-2z³
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z²
4z
5
z² bz 5
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
25
4z³
5z²
5z²
20z
-2z³
This shows that b must be -2
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z²
4z
5
z² -2z 5
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
25
4z³
5z²
5z²
20z
-2z³
This shows that b must be -2
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z²
4z
5
z² -2z 5
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
25
4z³
5z²
5z²
20z
-2z³
Now the last spaces in the table can be filled in
-8z²
-10z
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z²
4z
5
z² -2z 5
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25
z4
Factorising using a table
25
4z³
5z²
5z²
20z
-2z³
-8z²
-10z
and you can see that the term in z²is 2z² and the term in z is 10z, as they should be.
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)
Factorising by inspection
Now you can solve the equation by applying the quadratic formula to z²- 2z + 5 = 0.
The solutions of the equation are z = -2 + i, -2 - i, 1 + 2i, 1 – 2i.
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Factorising polynomials
Click here to see factorising by inspection
Click here to see this example of factorising using a table again
Click here to end the presentation
Click here to see polynomial division
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Algebraic long division
Divide z4 + 2z³+ 2z² + 10z + 25 by z² + 4z + 5
2 4 3 24 5 2 2 10 25z z z z z z
z² + 4z + 5 is the divisor
The quotient will be here.
z4 + 2z³ + 2z² + 10z + 25 is the dividend
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Algebraic long division
2 4 3 24 5 2 2 10 25z z z z z z
First divide the first term of the dividend, z4, by z² (the first term of the divisor).
This gives z². This will be the first term of the quotient.
z²
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Algebraic long division
2 4 3 24 5 2 2 10 25z z z z z z
z²
Now multiply z² by z² + 4z + 5
and subtract
z4 + 4z³ + 5z²
-2z³ - 3z²
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Algebraic long division
2 4 3 24 5 2 2 10 25z z z z z z
z²
z4 + 4z³ + 5z²
-2z³ - 3z²Bring down the next term, 10z
+ 10z
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Algebraic long division
2 4 3 24 5 2 2 10 25z z z z z z
z²
z4 + 4z³ + 5z²
-2z³ - 3z² + 10z
Now divide -2z³, the first term of -2z³ - 3z² + 5, by z², the first term of the divisor
which gives -2z
- 2z
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Algebraic long division
2 4 3 24 5 2 2 10 25z z z z z z
z²
z4 + 4z³ + 5z²
-2z³ - 3z² + 10z
- 2z
Multiply -2z by z² + 4z + 5
and subtract
-2z³- 8z²- 10z
5z²+ 20z
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Algebraic long division
2 4 3 24 5 2 2 10 25z z z z z z
z²
z4 + 4z³ + 5z²
-2z³ - 3z² + 10z
- 2z
-2z³- 8z²- 10z
5z²+ 20z
Bring down the next term, 25
+ 25
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Algebraic long division
2 4 3 24 5 2 2 10 25z z z z z z
z²
z4 + 4z³ + 5z²
-2z³ - 3z² + 10z
- 2z
-2z³- 8z²- 10z
5z²+ 20z + 25
Divide 5z², the first term of 5z² + 20z + 25, by z², the first term of the divisor which gives 5
+ 5
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Algebraic long division
2 4 3 24 5 2 2 10 25z z z z z z
z²
z4 + 4z³ + 5z²
-2z³ - 3z² + 10z
- 2z
-2z³- 8z²- 10z
5z²+ 20z + 25
+ 5
Multiply z² + 4z + 5 by 5 Subtracting gives 0 as there is no remainder.
5z² + 20z + 25 0
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z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)
Factorising by inspection
Now you can solve the equation by applying the quadratic formula to z²- 2z + 5 = 0.
The solutions of the equation are z = -2 + i, -2 - i, 1 + 2i, 1 – 2i.
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Factorising polynomials
Click here to see factorising by inspection
Click here to see factorising using a table
Click here to end the presentation
Click here to see this example of polynomial division again