Factorising quartics

The example in this presentation is from Example 2.10 in the FP1 textbook.

The aim is to factorise the quartic expression

z4 + 2z³ + 2z² + 10z + 25

into two quadratic factors, where one factor is z² + 4z + 5.

Factorising polynomials

This PowerPoint presentation demonstrates three methods of factorising a quartic into two quadratic factors when you know one quadratic factor.

Click here to see factorising by inspection

Click here to see factorising using a table

Click here to see polynomial division

Write the unknown quadratic as

az² + bz + c.

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(az² + bz + c)

Factorising by inspection

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(az² + bz + c)

Imagine multiplying out the brackets. The only way of getting a term in z4 is by multiplying z2 by az2, giving az4.

So a must be 1.

Factorising by inspection

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(1z² + bz + c)

Imagine multiplying out the brackets. The only way of getting a term in z4 is by multiplying z2 by az2, giving az4.

So a must be 1.

Factorising by inspection

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² + bz + c)

Now think about the constant term. You can only get a constant term by multiplying 5 by c, giving 5c.

So c must be 5.

Factorising by inspection

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² + bz + 5)

Now think about the constant term. You can only get a constant term by multiplying 5 by c, giving 5c.

So c must be 5.

Factorising by inspection

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² + bz + 5)

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

4z multiplied by 5 gives 20z

5 multiplied by bz gives 5bz

So 20z + 5bz = 10z

therefore b must be -2.

Factorising by inspection

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)

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

4z multiplied by 5 gives 20z

5 multiplied by bz gives 5bz

So 20z + 5bz = 10z

therefore b must be -2.

Factorising by inspection

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)

You can check by looking at the z² term. When you multiply out the brackets, you get three terms in z².

z² multiplied by 5 gives 5z²

4z multiplied by -2z gives -8z²5z² - 8z² + 5z² = 2z²

as it should be!

Factorising by inspection

5 multiplied by z² gives 5z²

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)

Factorising by inspection

Now you can solve the equation by applying the quadratic formula to z²- 2z + 5 = 0.

The solutions of the equation are z = -2 + j, -2 - j, 1 + 2j, 1 – 2j.

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

Write the unknown quadratic as az² + bz + c.

Factorising using a table

z²

4z

5

az² bz c

z²

4z

5

az² bz c

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

The only z4 term appears here,

so this must be z4.

z4

Factorising using a table

z²

4z

5

az² bz c

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

This means that a must be 1.

z²

4z

5

1z² bz c

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

This means that a must be 1.

z²

4z

5

z² bz c

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

The constant term, 25, must appear here

25

z²

4z

5

z² bz c

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

25

so c must be 5

z²

4z

5

z² bz 5

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

25

so c must be 5

z²

4z

5

z² bz 5

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

25

Four more spaces in the table can now be filled in

4z³

5z²

5z²

20z

z²

4z

5

z² bz 5

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

25

4z³

5z²

5z²

20z

This space must contain an z³ termand to make a total of 2z³, this must be -2z³

-2z³

z²

4z

5

z² bz 5

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

25

4z³

5z²

5z²

20z

-2z³

This shows that b must be -2

z²

4z

5

z² -2z 5

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

25

4z³

5z²

5z²

20z

-2z³

This shows that b must be -2

z²

4z

5

z² -2z 5

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

25

4z³

5z²

5z²

20z

-2z³

Now the last spaces in the table can be filled in

-8z²

-10z

z²

4z

5

z² -2z 5

The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25

z4

Factorising using a table

25

4z³

5z²

5z²

20z

-2z³

-8z²

-10z

and you can see that the term in z²is 2z² and the term in z is 10z, as they should be.

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)

Factorising by inspection

Now you can solve the equation by applying the quadratic formula to z²- 2z + 5 = 0.

The solutions of the equation are z = -2 + j, -2 - j, 1 + 2j, 1 – 2j.

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 z4 + 2z³+ 2z² + 10z + 25 by z² + 4z + 5

2 4 3 24 5 2 2 10 25z z z z z z

z² + 4z + 5 is the divisor

The quotient will be here.

z4 + 2z³ + 2z² + 10z + 25 is the dividend

Algebraic long division

2 4 3 24 5 2 2 10 25z z z z z z

First divide the first term of the dividend, z4, by z² (the first term of the divisor).

This gives z². This will be the first term of the quotient.

z²

Algebraic long division

2 4 3 24 5 2 2 10 25z z z z z z

z²

Now multiply z² by z² + 4z + 5

and subtract

z4 + 4z³ + 5z²

-2z³ - 3z²

Algebraic long division

2 4 3 24 5 2 2 10 25z z z z z z

z²

z4 + 4z³ + 5z²

-2z³ - 3z²Bring down the next term, 10z

+ 10z

Algebraic long division

2 4 3 24 5 2 2 10 25z z z z z z

z²

z4 + 4z³ + 5z²

-2z³ - 3z² + 10z

Now divide -2z³, the first term of -2z³ - 3z² + 5, by z², the first term of the divisor

which gives -2z

- 2z

Algebraic long division

2 4 3 24 5 2 2 10 25z z z z z z

z²

z4 + 4z³ + 5z²

-2z³ - 3z² + 10z

- 2z

Multiply -2z by z² + 4z + 5

and subtract

-2z³- 8z²- 10z

5z²+ 20z

Algebraic long division

2 4 3 24 5 2 2 10 25z z z z z z

z²

z4 + 4z³ + 5z²

-2z³ - 3z² + 10z

- 2z

-2z³- 8z²- 10z

5z²+ 20z

Bring down the next term, 25

+ 25

Algebraic long division

2 4 3 24 5 2 2 10 25z z z z z z

z²

z4 + 4z³ + 5z²

-2z³ - 3z² + 10z

- 2z

-2z³- 8z²- 10z

5z²+ 20z + 25

Divide 5z², the first term of 5z² + 20z + 25, by z², the first term of the divisor which gives 5

+ 5

Algebraic long division

2 4 3 24 5 2 2 10 25z z z z z z

z²

z4 + 4z³ + 5z²

-2z³ - 3z² + 10z

- 2z

-2z³- 8z²- 10z

5z²+ 20z + 25

+ 5

Multiply z² + 4z + 5 by 5 Subtracting gives 0 as there is no remainder.

5z² + 20z + 25 0

z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5)

Factorising by inspection

Now you can solve the equation by applying the quadratic formula to z²- 2z + 5 = 0.

The solutions of the equation are z = -2 + j, -2 - j, 1 + 2j, 1 – 2j.

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