Download presentation

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 division**

x² Now multiply x² by x - 3 and subtract

31
**Algebraic long division**

x² - 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 division**

x² + 2x - 4x Multiply 2x by x - 3 2x² - 6x 2x and subtract

34
**Algebraic long division**

x² + 2x - 6 - 4x Bring down the next term, -6 2x² - 6x 2x

35
**Algebraic long division**

x² + 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 division**

x² + 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

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google