Factorising quartics

Factorising quartics

One solution of the quartic equation

z4 + 2z³ + 2z² + 10z + 25 = 0

Solve the equation.

is z = -2 + i.


Since z = -2 + i is a solution,another solution is the complex conjugate z = -2 - i .Therefore two factors of the quartic expression are (z + 2 – i) and (z + 2 + i).

So a quadratic factor is (z + 2 – i)(z + 2 + i).

Multiplying out gives

(z + 2)² - (i)²

= z² + 4x + 4 - (-1)

= z² + 4x + 5


Now you need 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.


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.


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.


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.


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.


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.


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!


5 multiplied by z² gives 5z²

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


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

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


Click here to see this example of factorising by inspection again


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

Write the unknown quadratic as az² + bz + c.


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


z²

4z

5

az² bz c


z4


This means that a must be 1.

z²

4z

5

1z² bz c


z4


This means that a must be 1.

z²

4z

5

z² bz c


z4


The constant term, 25, must appear here

25

z²

4z

5

z² bz c


z4


25

so c must be 5

z²

4z

5

z² bz 5


z4


25

so c must be 5

z²

4z

5

z² bz 5


z4


25

Four more spaces in the table can now be filled in

4z³

5z²

5z²

20z

z²

4z

5

z² bz 5


z4


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


z4


25

4z³

5z²

5z²

20z

-2z³

This shows that b must be -2

z²

4z

5

z² -2z 5


z4


25

4z³

5z²

5z²

20z

-2z³

This shows that b must be -2

z²

4z

5

z² -2z 5


z4


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


z4


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)






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



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


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


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


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


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


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


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


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


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)








Click here to see this example of polynomial division again

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Factorising quartics