1. Factorising polynomials
This PowerPoint presentation demonstrates three methods of factorising a polynomial when you know one linear factor.
Click here to see factorising by inspection
Click here to see factorising using a table
Click here to see polynomial division

2. Factorising by inspection
If you divide x³ - x² - 4x – 6 (cubic) by x – 3 (linear), then the result must be quadratic.
Write the quadratic as ax² + bx + c.
x³ – x² – 4x - 6 = (x – 3)(ax² + bx + c)

3. Factorising by inspection
Imagine multiplying out the brackets. The only way of getting a term in x³ is by multiplying x by ax², giving ax³.
x³ – x² – 4x - 6 = (x – 3)(ax² + bx + c)
So a must be 1.

4. Factorising by inspection
Imagine multiplying out the brackets. The only way of getting a term in x³ is by multiplying x by ax², giving ax³.
x³ – x² – 4x - 6 = (x – 3)(1x² + bx + c)
So a must be 1.

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

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

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

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

9. Factorising by inspection
You can check by looking at the x term. When you multiply out the brackets, you get two terms in x.
-3 multiplied by 2x gives -6x
x³ – x² – 4x - 6 = (x – 3)(x² + 2x + 2)
x multiplied by 2 gives 2x
-6x + 2x = -4x as it should be!

10. Factorising by inspection
Now you can solve the equation by applying the quadratic formula to x²+ 2x + 2 = 0.
x³ – x² – 4x - 6 = (x – 3)(x² + 2x + 2)
The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.

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 see polynomial division
Click here to end the presentation

12. 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.

13. Factorising using a table
If you divide x³ - x² - 4x - 6 (cubic) by x – 3 (linear), then the result must be quadratic.
Write the quadratic as ax² + bx + c. x -3
ax² bx c

14. Factorising using a table
The result of multiplying out using this table has to be x³ - x² - 4x - 6 x -3
ax² bx c x³
The only x³ term appears here,
so this must be x³.

15. Factorising using a table
The result of multiplying out using this table has to be x³ - x² - 4x - 6 x -3
ax² bx c x³
This means that a must be 1.

16. Factorising using a table
The result of multiplying out using this table has to be x³ - x² - 4x - 6 x -3
1x² bx c x³
This means that a must be 1.

17. Factorising using a table
The result of multiplying out using this table has to be x³ - x² - 4x - 6 x -3
x² bx c x³ -6
The constant term, -6, must appear here

18. Factorising using a table
The result of multiplying out using this table has to be x³ - x² - 4x - 6 x -3
x² bx c x³ -6
so c must be 2

19. Factorising using a table
The result of multiplying out using this table has to be x³ - x² - 4x - 6 x -3
x² bx x³ -6
so c must be 2

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

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

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

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

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

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

26. Factorising by inspection
Now you can solve the equation by applying the quadratic formula to x²- 2x + 2 = 0.
x³ – x² – 4x - 6 = (x – 3)(x² - 2x + 2)
The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.

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 see polynomial division
Click here to end the presentation

28. Algebraic long division
Divide x³ - x² - 4x - 6 by x - 3
x - 3 is the divisor
x³ - x² - 4x - 6 is the dividend
The quotient will be here.

29. Algebraic long division
First divide the first term of the dividend, x³, by x (the first term of the divisor). x²
This gives x². This will be the first term of the quotient.

30. Algebraic long divisionx²
Now multiply x² by x - 3
and subtract

31. Algebraic long divisionx²
- 4x
Bring down the next term, -4x

32. Algebraic long division
Now divide 2x², the first term of 2x² - 4x, by x, the first term of the divisor x²
+ 2x
- 4x
which gives 2x

33. Algebraic long divisionx²
+ 2x
- 4x
Multiply 2x by x - 3
2x² - 6x 2x
and subtract

34. Algebraic long divisionx²
+ 2x
- 6
- 4x
Bring down the next term, -6
2x² - 6x 2x

35. Algebraic long divisionx²
+ 2x
+ 2
Divide 2x, the first term of 2x - 6, by x, the first term of the divisor
- 4x
2x² - 6x 2x
- 6
which gives 2

36. Algebraic long divisionx²
+ 2x
+ 2
- 4x
2x² - 6x
Multiply x - 3 by 2 2x
- 6
Subtracting gives 0 as there is no remainder.
2x - 6

37. Factorising by inspection
So x³ – x² – 4x - 6 = (x – 3)(x² - 2x + 2)
Now you can solve the equation by applying the quadratic formula to x²- 2x + 2 = 0.
The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.

38. 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