1. Factorising polynomials This PowerPoint presentation demonstrates two methods of factorising a polynomial when you know one factor (perhaps by using the factor theorem). Click here to see factorising by inspection Click here to see factorising using a table

2. If you divide 2x³ - 5x² - 4x – 3 (cubic) by x – 3 (linear), then the result must be quadratic. Write the quadratic as ax² + bx + c. 2x³ – 5x² – 4x + 3 = (x – 3)(ax² + bx + c) Factorising by inspection

3. Imagine multiplying out the brackets. The only way of getting a term in x³ is by multiplying x by ax², giving ax³. So a must be 2. Factorising by inspection 2x³ – 5x² – 4x + 3 = (x – 3)(ax² + bx + c)

4. Imagine multiplying out the brackets. The only way of getting a term in x³ is by multiplying x by ax², giving ax³. So a must be 2. Factorising by inspection 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx + c)

5. Now think about the constant term. You can only get a constant term by multiplying –3 by c, giving –3c. So c must be Factorising by inspection 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx + c)

6. Now think about the constant term. You can only get a constant term by multiplying –3 by c, giving –3c. So c must be -1. Factorising by inspection 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx - 1)

7. Now think about the x²x² term. When you multiply out the brackets, you get two x² terms. -3 multiplied by 2x² gives –6x² x multiplied by bx gives bx² So –6x² + bx² = -5x² therefore b must be 1. 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx - 1) Factorising by inspection

8. Now think about the x² term. When you multiply out the brackets, you get two x² terms. -3 multiplied by 2x² gives –6x² x multiplied by bx gives bx² So –6x² + bx² = -5x² therefore b must be 1. 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + 1x - 1) Factorising by inspection

9. You can check by looking at the x term. When you multiply out the brackets, you get two terms in x.x. -3 multiplied by x gives –3x x multiplied by –1 gives -x-x -3x – x = -4x as it should be! 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1) Factorising by inspection

10. 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1) Factorising by inspection Now factorise the quadratic in the usual way. = (x (x – 3)(2x – 1)(x + 1)

11. 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

12. 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

13. If you divide 2x³ - 5x² - 4x 4x + 3 (cubic) by x – 3 (linear), then the result must be quadratic. Write the quadratic as ax² + bx + c. x -3 ax² bx c Factorising using a table

14. x -3 ax² bx c The result of multiplying out using this table has to be 2x³ - 5x² - 4x 4x + 3 The only x³ term appears here, so this must be 2x³. 2x³2x³ Factorising using a table

15. This means that a must be 2. x -3 ax² bx c 2x³2x³ The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 Factorising using a table

16. This means that a must be 2. x -3 2x² bx c 2x³2x³ The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 Factorising using a table

17. The constant term, 3, must appear here 3 x -3 2x² bx c 2x³2x³ The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 Factorising using a table

18. so c must be –1. 3 x -3 2x² bx c 2x³2x³ The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 Factorising using a table

19. so c must be –1. 3 x -3 2x² bx -1 2x³2x³ The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 Factorising using a table

20. 3 x -3 2x² bx -1 2x³2x³ Two more spaces in the table can now be filled in -6x² -x-x The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 Factorising using a table

21. This space must contain an x² term and to make a total of –5x², this must be x²x² x²x² 3 x -3 2x² bx -1 2x³2x³ -6x² -x-x The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 Factorising using a table

22. This shows that b must be 1. x²x² 3 x -3 2x² bx -1 2x³2x³ The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 Factorising using a table -6x² -x-x

23. This shows that b must be 1. x²x² 3 x -3 2x² 1x -1 2x³2x³ The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 Factorising using a table -6x² -x-x

24. Now the last space in the table can be filled in The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 x²x² -3x 3 x -3 2x² x -1 2x³2x³ -6x² -x-x Factorising using a table

25. and you can see that the term in x is –4x, as it should be. So 2x³ - 5x² - 4x 4x + 3 = (x (x – 3)(2x² + x – 1) x²x² 3 x -3 2x² x -1 2x³2x³ The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3 -6x² -x-x Factorising using a table -3x

26. 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1) Factorising by inspection Now factorise the quadratic in the usual way. = (x (x – 3)(2x – 1)(x + 1)

27. Factorising polynomials Click here to see this example of factorising using a table again Click here to see factorising by inspection Click here to end the presentation