# Factorising polynomials - PowerPoint PPT Presentation

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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. Factorising by inspection.

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Factorising polynomials

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### Factorising polynomials

This PowerPoint presentation demonstrates two methods of factorising a polynomial when you know one factor (perhaps by using the factor theorem).

### 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)

### 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.

### 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.

### 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.

### 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.

### 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.

### 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.

### 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!

### 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)

x² -3x - 4

2x

3

### 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:

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.

ax² bxc

x

-3

### 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.

### Factorising using a table

ax² bxc

x

-3

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

2x³

The only x³ term appears here,

so this must be 2x³.

### Factorising using a table

ax² bxc

x

-3

2x³

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

This means that a must be 2.

### Factorising using a table

2x² bxc

x

-3

2x³

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

This means that a must be 2.

### Factorising using a table

2x² bxc

x

-3

2x³

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

3

The constant term, 3, must appear here

### Factorising using a table

2x² bxc

x

-3

2x³

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

3

so c must be –1.

### Factorising using a table

2x² bx-1

x

-3

2x³

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

3

so c must be –1.

### Factorising using a table

2x² bx -1

x

-3

2x³

3

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

-x

-6x²

Two more spaces in the table can now be filled in

### Factorising using a table

2x² bx -1

x

-3

2x³

3

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

-x

-6x²

This space must contain an x² term

and to make a total of –5x², this must be x²

### Factorising using a table

2x² bx -1

x

-3

2x³

3

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

-x

-6x²

This shows that b must be 1.

### Factorising using a table

2x² 1x -1

x

-3

2x³

3

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

-x

-6x²

This shows that b must be 1.

### Factorising using a table

2x² x -1

x

-3

2x³

-x

-6x²

3

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

-3x

Now the last space in the table can be filled in

### Factorising using a table

2x² x -1

x

-3

2x³

3

The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3

-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)

### 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)

### Factorising polynomials

Click here to see this example of factorising using a table again