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

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.

So a must be 2.

Factorising by inspection

2x³ – 5x² – 4x + 3 = (x – 3)(ax² + bx + c)

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)

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 + c)

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)

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 = -5x²therefore b must be 1.

2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx - 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

x multiplied by bx gives bx²

So –6x² + bx² = -5x²therefore b must be 1.

2x³ – 5x² – 4x + 3 = (x – 3)(2x² + 1x - 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

x multiplied by –1 gives -x

-3x – x = -4x as it should be!

2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1)

Factorising by inspection

2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1)

Factorising by inspection

Now factorise the quadratic in the usual way.

= (x – 3)(2x – 1)(x + 1)

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

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 -3x - 4

2x³ -6x² -8x

3x² -9x -12

So (2x + 3)(x² - 3x – 4) = 2x³ - 3x² - 17x - 12The method you are going to see now is basically the reverse of this process.

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

Factorising using a table

x

-3

ax² bx c

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

The only x³ term appears here,

so this must be 2x³.

2x³

Factorising using a table

This means that a must be 2.

x

-3

ax² bx c

2x³

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

Factorising using a table

This means that a must be 2.

x

-3

2x² bx c

2x³

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

Factorising using a table

The constant term, 3, must appear here

3

x

-3

2x² bx c

2x³

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

Factorising using a table

so c must be –1.

3

x

-3

2x² bx c

2x³

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

Factorising using a table

so c must be –1.

3

x

-3

2x² bx -12x³

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

Factorising using a table

3

x

-3

2x² bx -12x³

Two more spaces in the table can now be filled in

-6x²

-x

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

Factorising using a table

This space must contain an x² term

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

x²

3

x

-3

2x² bx -12x³

-6x²

-x

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

Factorising using a table

This shows that b must be 1.

x²

3

x

-3

2x² bx -12x³

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

Factorising using a table

-6x²

-x

This shows that b must be 1.

x²

3

x

-3

2x² 1x -12x³

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

Factorising using a table

-6x²

-x

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

x²

-3x 3

x

-3

2x² x -1

2x³

-6x²

-x

Factorising using a table

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)

x²

3

x

-3

2x² x -1

2x³

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

-6x²

-x

Factorising using a table

-3x

2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1)

Factorising by inspection

Now factorise the quadratic in the usual way.

= (x – 3)(2x – 1)(x + 1)

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