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

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)

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 1.

Factorising by inspection

x³ – x² – 4x - 6 = (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 1.

Factorising by inspection

x³ – x² – 4x - 6 = (x – 3)(1x² + 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 2.

Factorising by inspection

x³ – x² – 4x - 6 = (x – 3)(x² + 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 2.

Factorising by inspection

x³ – x² – 4x - 6 = (x – 3)(x² + bx + 2)

Now think about the x² term. When you multiply out the brackets, you get two x² terms.

-3 multiplied by x² gives –3x²

x multiplied by bx gives bx²

So –3x² + bx² = -1x²therefore b must be 2.

x³ – x² – 4x - 6 = (x – 3)(x² + bx + 2)

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

x multiplied by bx gives bx²

So –3x² + bx² = -1x²therefore b must be 2.

x³ – x² – 4x - 6 = (x – 3)(x² + 2x + 2)

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

x multiplied by 2 gives 2x

-6x + 2x = -4x as it should be!

x³ – x² – 4x - 6 = (x – 3)(x² + 2x + 2)

Factorising by inspection

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.

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

Click here to see polynomial division

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 x³ - x² - 4x - 6 (cubic) by x – 3 (linear), then the result must be quadratic.Write the quadratic as ax² + bx + c.

Factorising using a table

x

-3

ax² bx c

x

-3

ax² bx c

The result of multiplying out using this table has to be x³ - x² - 4x - 6

The only x³ term appears here,

so this must be x³.

x³

Factorising using a table

x

-3

ax² bx c

The result of multiplying out using this table has to be x³ - x² - 4x - 6

This means that a must be 1.

x³

Factorising using a table

x

-3

1x² bx c

The result of multiplying out using this table has to be x³ - x² - 4x - 6

This means that a must be 1.

x³

Factorising using a table

x

-3

x² bx c

The result of multiplying out using this table has to be x³ - x² - 4x - 6

The constant term, -6, must appear here

x³

Factorising using a table

-6

x

-3

x² bx c

The result of multiplying out using this table has to be x³ - x² - 4x - 6

so c must be 2

x³

Factorising using a table

-6

x

-3

x² bx 2

The result of multiplying out using this table has to be x³ - x² - 4x - 6

so c must be 2

x³

Factorising using a table

-6

x

-3

x² bx 2

The result of multiplying out using this table has to be x³ - x² - 4x - 6

Two more spaces in the table can now be filled in

x³

Factorising using a table

-6

2x

-3x²

x

-3

x² bx 2

The result of multiplying out using this table has to be x³ - x² - 4x - 6

This space must contain an x² term

x³

Factorising using a table

-6

2x

-3x²

and to make a total of –x², this must be 2x²

2x²

x

-3

x² bx 2

The result of multiplying out using this table has to be x³ - x² - 4x - 6

This shows that b must be 2

x³

Factorising using a table

-6

2x

-3x²

2x²

x

-3

x² 2x 2

The result of multiplying out using this table has to be x³ - x² - 4x - 6

This shows that b must be 2

x³

Factorising using a table

-6

2x

-3x²

2x²

x

-3

x² 2x 2

The result of multiplying out using this table has to be x³ - x² - 4x - 6

Now the last space in the table can be filled in

x³

Factorising using a table

-6

2x

-3x²

2x²

-6x

x

-3

x² 2x 2

The result of multiplying out using this table has to be x³ - x² - 4x - 6

and you can see that the term in x is -4x, as it should be.

x³

Factorising using a table

-6

2x

-3x²

2x²

-6x

So x³ - x² - 4x - 6 = (x – 3)(x² + 2x + 2)

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.

Factorising polynomials

Click here to see factorising by inspection

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

Click here to end the presentation

Click here to see polynomial division

Algebraic long division

Divide x³ - x² - 4x - 6 by x - 3

3 23 4 6x x x xx - 3 is the divisor

The quotient will be here.

x³ - x² - 4x - 6 is the dividend

Algebraic long division

3 23 4 6x x x x

First divide the first term of the dividend, x³, by x (the first term of the divisor).

This gives x². This will be the first term of the quotient.

x²

Algebraic long division

3 23x x

3 23 4 6x x x xx²

Now multiply x² by x - 3

and subtract

22x

Algebraic long division

3 23x x

3 23 4 6x x x xx²

22xBring down the next term, -4x

- 4x

Algebraic long division

3 23x x

3 23 4 6x x x xx²

22x - 4x

Now divide 2x², the first term of 2x² - 4x, by x, the first term of the divisor

which gives 2x

+ 2x

Algebraic long division

3 23x x

3 23 4 6x x x xx²

22x - 4x

+ 2x

Multiply 2x by x - 3

and subtract

2x² - 6x

2x

Algebraic long division

3 23x x

3 23 4 6x x x xx²

22x - 4x

+ 2x

2x² - 6x

2x

Bring down the next term, -6

- 6

Algebraic long division

3 23x x

3 23 4 6x x x xx²

22x - 4x

+ 2x

2x² - 6x

2x - 6

Divide 2x, the first term of 2x - 6, by x, the first term of the divisor which gives 2

+ 2

Algebraic long division

3 23x x

3 23 4 6x x x xx²

22x - 4x

+ 2x

2x² - 6x

2x - 6

+ 2

Multiply x - 3 by 2

Subtracting gives 0 as there is no remainder.

2x - 60

Factorising by inspection

Now you can solve the equation by applying the quadratic formula to x²- 2x + 2 = 0.

So x³ – x² – 4x - 6 = (x – 3)(x² - 2x + 2)

The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.

Factorising polynomials

Click here to see factorising by inspection

Click here to see factorising using a table

Click here to end the presentation

Click here to see this example of polynomial division again