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 i).
So a quadratic factor is (z + 2 – i)(z i).
Multiplying out gives
(z + 2)² - (i)²
= z² + 4x (-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: 2x 3
x² x
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. z² 4z 5
az² bz c

17. Factorising using a table
The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5
az² bz c 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 z² 4z 5
az² bz c 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 z² 4z 5
1z² bz c 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 z² 4z 5
z² bz c 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 z² 4z 5
z² bz c 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 z² 4z 5
z² bz 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 z² 4z 5
z² bz z4
5z²
4z³
20z
5z² 25
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 z² 4z 5
z² bz z4
-2z³
5z²
4z³
20z
5z² 25
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 z² 4z 5
z² bz z4
-2z³
5z²
4z³
20z
5z² 25
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 z² 4z 5
z² z z4
-2z³
5z²
4z³
20z
5z² 25
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 z² 4z 5
z² z z4
-2z³
5z²
4z³
-8z²
20z
5z²
-10z 25
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 z² 4z 5
z² z z4
-2z³
5z²
4z³
-8z²
20z
5z²
-10z 25
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 divisionz²
z4 + 4z³ + 5z²
Now multiply z² by z² + 4z + 5
-2z³ - 3z²
and subtract

34. Algebraic long divisionz²
+ 10z
z4 + 4z³ + 5z²
-2z³ - 3z²
Bring down the next term, 10z

35. Algebraic long divisionz²
- 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 divisionz²
- 2z
z4 + 4z³ + 5z²
-2z³ - 3z²
+ 10z
Multiply -2z by z² + 4z + 5
-2z³- 8z²- 10z
5z²+ 20z
and subtract

37. Algebraic long divisionz²
- 2z
+ 25
z4 + 4z³ + 5z²
-2z³ - 3z²
+ 10z
Bring down the next term, 25
-2z³- 8z²- 10z
5z²+ 20z

38. Algebraic long divisionz²
- 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 divisionz²
- 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

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