1. Factorising quartics The example in this presentation is from Example 2.10 in the FP1 textbook. The aim is to factorise the quartic expression z 4 + 2z³ + 2z² + 10z + 25 into two quadratic factors, where one factor is z² + 4z + 5.

2. 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

3. Write the unknown quadratic as az² + bz + c. z 4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(az² + bz + c) Factorising by inspection

4. z 4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(az² + bz + c) Imagine multiplying out the brackets. The only way of getting a term in z4 z4 is by multiplying z2 z2 by az 2, giving az 4. So a must be 1. Factorising by inspection

5. z 4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(1z² + bz + c) Imagine multiplying out the brackets. The only way of getting a term in z 4 is by multiplying z 2 by az 2, giving az 4. So a must be 1. Factorising by inspection

6. z 4 + 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.5c. So c must be 5. Factorising by inspection

7. z 4 + 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

8. z 4 + 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.z. 4z 4z multiplied by 5 gives 20z 5 multiplied by bz gives 5bz So 20z + 5bz = 10z therefore b must be -2. Factorising by inspection

9. z 4 + 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

10. z 4 + 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 5z²5z² 4z 4z multiplied by -2z gives -8z² 5z² - 8z² + 5z² = 2z²2z² as it should be! Factorising by inspection 5 multiplied by z² gives 5z²5z²

11. z 4 + 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 + j, -2 - j, 1 + 2j, 1 – 2j.

12. 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

13. 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: 2x32x3 x² -3x - 4 2x³2x³-6x²-8x 3x²3x²-9x-12 So (2x + 3)(x² - 3x 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

14. Write the unknown quadratic as az² + bz + c. Factorising using a table z²4z5z²4z5 az² bz c

15. z²4z5z²4z5 The result of multiplying out using this table has to be z4 z4 + 2z³ + 2z² + 10z + 25 The only z4 z4 term appears here, so this must be z4.z4. z4z4 Factorising using a table

16. z²4z5z²4z5 az² bz c The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table This means that a must be 1.

17. z²4z5z²4z5 1z² bz c The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table This means that a must be 1.

18. z²4z5z²4z5 z² bz c The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table The constant term, 25, must appear here 25

19. z²4z5z²4z5 z² bz c The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table 25 so c must be 5

20. z²4z5z²4z5 z² bz 5 The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table 25 so c must be 5

21. z²4z5z²4z5 z² bz 5 The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table 25 Four more spaces in the table can now be filled in 4z³4z³ 5z²5z² 5z²5z² 20z

22. z²4z5z²4z5 z² bz 5 The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table 25 4z³4z³ 5z²5z² 5z²5z² 20z This space must contain an z³ term and to make a total of 2z³, this must be -2z³

23. z²4z5z²4z5 z² bz 5 The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table 25 4z³4z³ 5z²5z² 5z²5z² 20z -2z³ This shows that b must be -2

24. z²4z5z²4z5 z² -2z 5 The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table 25 4z³4z³ 5z²5z² 5z²5z² 20z -2z³ This shows that b must be -2

25. z²4z5z²4z5 z² -2z 5 The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table 25 4z³4z³ 5z²5z² 5z²5z² 20z -2z³ Now the last spaces in the table can be filled in -8z² -10z

26. z²4z5z²4z5 z² -2z 5 The result of multiplying out using this table has to be z 4 + 2z³ + 2z² + 10z + 25 z4z4 Factorising using a table 25 4z³4z³ 5z²5z² 5z²5z² 20z -2z³ -8z² -10z and you can see that the term in z²is 2z²2z² and the term in z is 10z, as they should be.

27. z 4 + 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 + j, -2 - j, 1 + 2j, 1 – 2j.

28. 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

29. Algebraic long division Divide z4 z4 + 2z³+ 2z² + 10z + 25 by z² + 4z 4z + 5 + 4z 4z + 5 is the divisor The quotient will be here. z4 z4 + 2z³ + 2z² + 10z + 25 is the dividend

30. Algebraic long division First divide the first term of the dividend, z4,z4, by z² (the first term of the divisor). This gives z². This will be the first term of the quotient. z²z²

31. Algebraic long division z²z² Now multiply z²z² by z² + 4z + 5 and subtract z 4 + 4z³ + 5z² -2z³ - 3z²

32. Algebraic long division z²z² z 4 + 4z³ + 5z² -2z³ - 3z² Bring down the next term, 10z + 10z

33. Algebraic long division z²z² z 4 + 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

34. Algebraic long division z²z² z 4 + 4z³ + 5z² -2z³ - 3z²+ 10z - 2z Multiply -2z by z² + 4z 4z + 5 and subtract -2z³- 8z²- 10z 5z²+ 20z

35. Algebraic long division z²z² z 4 + 4z³ + 5z² -2z³ - 3z²+ 10z - 2z -2z³- 8z²- 10z 5z²+ 20z Bring down the next term, 25 + 25

36. Algebraic long division z²z² z 4 + 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

37. Algebraic long division z²z² z 4 + 4z³ + 5z² -2z³ - 3z²+ 10z - 2z -2z³- 8z²- 10z 5z²+ 20z+ 25 + 5 Multiply z² + 4z 4z + 5 by 5 Subtracting gives 0 as there is no remainder. 5z² + 20z + 25 0

38. z 4 + 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 + j, -2 - j, 1 + 2j, 1 – 2j.

39. 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