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. Factorising by inspection
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)

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.
2x³ – 5x² – 4x + 3 = (x – 3)(ax² + bx + c)
So a must be 2.

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.
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx + c)
So a must be 2.

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

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

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

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

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 x gives –3x
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1)
x multiplied by –1 gives -x
-3x – x = -4x as it should be!

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

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

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

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

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

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

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

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

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

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

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

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