Products and factors  (Page 3/4)

Let us consider a quadratic that is of the form $a{x}^2+bx$ . We can see here that $x$ is a common factor of both terms. Therefore, $a{x}^2+bx$ factorises to $x(ax+b)$ . For example, $8y^2+4y$ factorises to $4y(2y+1)$ .

Another type of quadratic is made up of the difference of squares. We know that:

This is true for any values of $a$ and $b$ , and more importantly since it is an equality, we can also write:

This means that if we ever come across a quadratic that is made up of a difference of squares, we can immediately write down what the factors are.

These types of quadratics are very simple to factorise. However, many quadratics do not fall into these categories and we need a more general method to factorise quadratics like $x^2-x-2$ ?

We can learn about how to factorise quadratics by looking at how two binomials are multiplied to get a quadratic. For example, $(x+2)(x+3)$ is multiplied out as:

We see that the $x^2$ term in the quadratic is the product of the $x$ -terms in each bracket. Similarly, the 6 in the quadratic is the product of the 2 and 3 in the brackets. Finally, the middle term is the sum of two terms.

So, how do we use this information to factorise the quadratic?

Let us start with factorising $x^2+5x+6$ and see if we can decide upon some general rules. Firstly, write down two brackets with an $x$ in each bracket and space for the remaining terms.

Next, decide upon the factors of 6. Since the 6 is positive, these are:

Therefore, we have four possibilities:

Next, we expand each set of brackets to see which option gives us the correct middle term.

We see that Option 3 (x+2)(x+3) is the correct solution. As you have seen that the process of factorising a quadratic is mostly trial and error, there is some information that can be used to simplify the process.

Method: factorising a quadratic

1. First, divide the entire equation by any common factor of the coefficients so as to obtain an equation of the form $a{x}^2+bx+c=0$ where $a$ , $b$ and $c$ have no common factors and $a$ is positive.
2. Write down two brackets with an $x$ in each bracket and space for the remaining terms.
   $$\left(x\right)\left(x\right)$$
3. Write down a set of factors for $a$ and $c$ .
4. Write down a set of options for the possible factors for the quadratic using the factors of $a$ and $c$ .
5. Expand all options to see which one gives you the correct answer.

There are some tips that you can keep in mind:

• If $c$ is positive, then the factors of $c$ must be either both positive or both negative. The factors are both negative if $b$ is negative, and are both positive if $b$ is positive. If $c$ is negative, it means only one of the factors of $c$ is negative, the other one being positive.
• Once you get an answer, multiply out your brackets again just to make sure it really works.