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Factorization

Quadratic Functions

Generally speaking, factorisation is the reverse of multiplying out. One important factorisation process is the reverse of multiplications such as this:

(x − 5)(x + 3) = x2 − 2x − 15.

Here we have muliplied out two linear factors to obtain a quadratic expression by using the distributive law.

To factorise a quadratic we have to go from an expression such as x2 − 2x − 15 to the linear factors (x − 5)(x + 3) which generate it when multiplied out. How can we do this?

The clue lies in the solutions of the equation x2 − 2x − 15 = 0 (called a quadratic equation). If we factorise the quadratic, the equation can be written as (x − 5)(x + 3) = 0. But a product of two factors can only be equal to zero if one or the other factor is equal to zero. So for the equation to hold, either x − 5 must be zero or x + 3 must be zero. Therefore the two possible solutions of the equation are x = 5 and x = −3. Looking at it the other way around, if we knew the solutions of the equation, we could find the factors: they just take the form (x − (solution)). In this example, the two solutions are 5 and −3, and so the two factors are (x5) and (x − (−3)) which simplifies to (x − 5) and (x + 3).

So it looks as though if we could solve quadratic equations, we could factorise quadratics. But there is a formula for solving quadratic equations which you have probably seen before: it says that

The solutions of the equation  Ax2 + Bx + C = 0
are given by the formula

For example, the quadratic x2 − 2x − 15 has A = 1, B = −2 and C = −15. So according to the formula the solutions of the equation x2 − 2x − 15 = 0 are:

as before.

Knowing the solutions, we can then write down the factors as explained earlier. Here then is the strategy to use:

Example

 

x2 − 12x + 35

Strategy

A = 1, B = −12, C = 35

Obtain the values of A, B and C by inspection.


Substitute the values into the quadratic formula and simplify.

x2 − 12x + 35 = (x − 5)(x − 7)

Write down the factors.

Try a few practice problems:

Exercise

Enter the values of A, B and C for the expression

6x2 − 18x + 43

A = B = C =

 

Enter the values of A, B and C for the expression

x2 + x − 6

A = B = C =

The correct values will be substituted below.

x =
 √( 2 − 4 )  
=
√(


2
x =
2xxxxx
=
2xxxx


 

x = or x =

Now see if you can write the factors:

x2 + x − 6 =

 

 

There are some complications that crop up when using this method to factorise a quadratic.

Firstly, the formula sometimes (depending on the values of the coefficients) involves the square root of a negative number.

Example

 

2x2 + 3x + 5

Strategy

A = 2, B = 3, C = 5

Obtain the values of A, B and C by inspection.



Substitute the values into the quadratic formula and simplify.

Square root of a negative number indicates the quadratic cannot be factorised using real numbers.

But −31 does not have a real square root, so this quadratic cannot be factorised using real numbers.

Now try a few exercises:

Exercise

Factorise each expression wherever possible or type "Impossible" if it cannot be factorised using real numbers.

1. x2 − 7x + 6 =

2. x2 + 7x + 6 =

3. x2 − 2x + 6 =

4. x2 − 11x + 30 =

5. x2 − 12x −28 =

 

The next complication is that sometimes there is only one solution to the formula.

Example

 

x2 − 8x + 16

Strategy

A = 1, B = −8, C =16

Obtain the values of A, B and C by inspection


Substitute the values into the quadratic formula and simplify.

Square root of 0 indicates the quadratic is a perfect square.

x2 − 8x + 16 = (x − 4)(x − 4) = (x − 4)2

Write down the perfect square.

Still more exercises for practice:

Exercise

Factorise each expression wherever possible or type "Impossible". Type two identical factors for perfect squares, (x+b)(x+b), or use the Excel and MATLAB notation (x+b)^2

1. x2 − 12x + 36 =

2. x2 + 10x + 16 =

3. x2 − 2x − 8 =

4. x2 − 11x + 42 =

5. x2 − 6x + 9 =

The third and final complication occurs for quadratic expressions where A is not equal to 1.

Example  
4x2 + 12x + 5
Strategy

A = 4, B = 12, C =5

Obtain the values of A, B and C by inspection

Substitute the values into the quadratic formula and simplify.

 

4x2 + 12x + 5 = 4(x + )(x + )

4x2 + 12x + 5 = 2(x + )2(x + )

4x2 + 12x + 5 = (2x + 5)(2x + 1)

To make the factorisation correct we have to multiply by the coefficient of x2 in the original quadratic, that is multiply by A. (No action is required if A = 1.)

Simplify if possible

Now you are fully armed to tackle any quadratic expression. Try these quadratic exercises.

Exercise

Factorise each expression wherever possible.

1. 2x2 − 5x − 3

2. 3x2 + x − 2

3. 4x2 + 4x − 3

4. 5x2 − 8x + 13

5. 5 − 9x − 2x2

6. 3 − 18x + 27x2

Exercise

Factorise this quadratic function:
x2 + x  + 

Working space:

Answer:

 

<< Exponential Functions | Factorisation Index | Quadratic Functions Factoriser >>

<< Quadratic Formula | Quadratic Functions | Quadratic Functions Factoriser >>

 

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