### Factorising quadratics when $a \ne 1$

The general quadratic form is $ax^2 \pm bx \pm c$ where $a, b, c \in\Bbb{R}$ ($a, b, c$ are in the set of real numbers). Up to this point, we have only attempted to factorise quadratics where $a=1$. In this section, we will look at quadratics where $a$ is other than $1$.

The methods available to us are many and various. I have a particular favourite, as it enabled me, slow as I am, to factorise these things in my head as easily as when $a=1$. However, I will cover several methods and you should use the one that works best for you.

The worked examples are different here as they work with the same quadratics, but using different methods. Hopefully, this will help you to pick the one that is best for you.

### Worked example 1: Factorising a quadratic with a constant factor

- Notes
- Factorise fully the quadratic expression $3x^2-3x-18$
This is an important type of problem to be aware of, as it can save you a lot of time in the long run, if you can recognise them to begin with.

- \[\begin{aligned} &3x^2-3x-18\\ &=3(x^2-x-6)\\ &=3(x-3)(x+2) \end{aligned}\]
This is question type which can cause people problem in not attacked from the right angle.

Note that the three terms making up this expression all have a factor of $3$.

Our first job is to factorise out the $3$.

Then we are left with a simple $a=1$ quadratic to solve.

- \[3(x-3)(x+2)\]

### Worked example 2: Factorise $3x^2 - 5x - 12$ (Group factorisation method)

- Notes
- Factorise completely the quadratic $3x^2 - 5x - 12$
This was the method I was taught at school and it has merit, though it's not my preferred method.

Firstly, you should notice that this is a

**one of each signs**example, as the constant is $-ve$.Advantages of this method is that it works fine and done correctly will always produce the correct answer.

Disadvantages are that it requires several factorisation steps, it is easy to make an error and it is quite a length process.

- $\begin{aligned} &3x^2 - 5x - 12 \hspace {50pt} \big(1 \big) \\ &= 3x^2-9x+4x-12 \hspace {14pt} \big(2 \big) \\ &=3x \big(x-3 \big) +4 \big(x-3 \big) \hspace {6pt} \big(3 \big) \\ &=\big(x-3 \big) \big(3x+4 \big) \hspace {29pt} \big(4 \big) \\ \end{aligned} $
- Multiply $a$ by $c$, including their signs. $3\big(-12\big) = -36$.
- We are now going to search for a pair of numbers which multiply to give $-36$ and add together to give $-5$. (It is worth noting at this point, that $a=1 \Rightarrow ac = c$ ($a \ is \ 1$ implies that $ac \ is \ c$).
- Write down the factor pairs of 36. $\begin{aligned} &1 \times 36\\ &2 \times 18\\ &3 \times 12\\ &4 \times 9\\ &6 \times 6 \end{aligned}$
- The only way to get a sum of -5 with any of these factor pairs is to use $4$ and $-9$
- Rewrite the quadratic with the middle term split according to the numbers we have found. (line 3)
- Do a linear factorisation on the successive pairs of terms. (line 3)
- Now we can factorise this new expression. Note that the term $\big(x-3 \big)$ is a common factor of the terms $3x\big(x-3 \big)$ and $+4\big(x-3 \big)$. (line 4)
- If you end up with a different expression in the two bracket pair on line 3, you have made an error, so don't try to struggle onwards if that happens.

- $\big(x-3 \big) \big(3x+4 \big)$
It is always a good idea to multiply out the factors, to make sure you end up with the quadratic you started with.

### Worked example 3: Factorise $3x^2 - 5x - 12$ (Magic factorisation method)

- Notes
- Factorise completely the quadratic $3x^2 - 5x - 12$
- $\begin{aligned} &3x^2 - 5x - 12 \hspace {50pt} \big(1 \big) \\ &=\big(3x-9 \big) \big(3x+4 \big) \div 3 \hspace {6pt} \big(2\big) \\ &=\big(x-3 \big) \big(3x+4 \big) \hspace {29pt} \big(3 \big) \\ \end{aligned} $
- Multiply $a$ by $c$, including their signs. $3\big(-12\big) = -36$.
- We are now going to search for a pair of numbers which multiply to give $-36$ and add together to give $-5$. (It is worth noting at this point, that $a=1 \Rightarrow ac = c$ ($a \ is \ 1$ implies that $ac \ is \ c$).
- Write down the factor pairs of 36. $\begin{aligned} &1 \times 36\\ &2 \times 18\\ &3 \times 12\\ &4 \times 9\\ &6 \times 6 \end{aligned}$
- The only way to get a sum of -5 with any of these factor pairs is to use $4$ and $-9$
- So far, so much like the last method. From here we diverge.
- Write down two pairs of brackets, with $3x$ at the beginning of both and the $-9$ and $+4$ finishing off both brackets.
- Clearly this is wrong, as $3x \times 3x = 9x^2$ so it looks as through it is going to be $3 \times$ too big.
- Divide one or both of the factors by $3$. $\big(3x-9\big)$ is divisible by 3, giving the correct answer.

- $\big(x-3 \big) \big(3x+4 \big)$
There are occasions when you will need to divide both brackets, eg one of by $2$ and the other by $3$, when your fake factorisation is $6 \times$ as big as it should be.

Look at the later Worked examples for more examples where this occurs.

From now on, I shall be only using this method, as I am lazy and it is easier.

### Worked example 4: Factorise $6x^2 + 17x + 12$ (Magic factorisation method)

- Notes
- Factorise completely the expression $6x^2 + 17x + 12$
- \[ \begin{aligned} &6x^2 + 17x + 12\\ &=(6x+8)(6x+9) \div 6 \\ &=(3x+4)(2x+3) \end{aligned}\]
Find a pair of numbers which have a sum $+17$ and product of $+72$

$+8$ and $+9$ fit the bill nicely.

We need to divide by $6$, but neither of the brackets has a factor of $6$.

However, the first will divide by $2$ and the 2nd has a factor of $3$. Dividing by $2$ and then $3$ has precisely the same as dividing by $6$

- $(3x+4)(2x+3)$

### Exercise 1: Factorising quadratics $a>1$, all signs $+$

**Instructions:**Factorise the following quadratic expressions- Q1\[2f^2+3f+1\]\[(2f+1)(f+1)\]
- Q2\[3a^2+4a+1\]\[(3a+1)(a+1)\]
- Q3\[5m^2+6m+1\]\[(5m+1)(m+1)\]
- Q4\[6s^2+5s+1\]\[(2s+1)(3s+1)\]
- Q5\[10g^2+7g+1\]\[(2g+1)(5g+1)\]
- Q6\[8w^2+6w+1\]\[(4w+1)(2w+1)\]
- Q7\[3z^2+10z+3\]\[(3z+1)(z+3)\]
- Q8\[6n^2+11n+3\]\[(3n+1)(2n+3)\]
- Q9\[6b^2+13b+6\]\[(2b+3)(3b+2)\]
- Q10\[4t^2+7t+3\]\[(t+1)(4t+3)\]

### Exercise 2: Factorising quadratics $a>1$, middle term $-$

**Instructions:**Factorise the following quadratic expressions- Q1\[2e^2-3e+1\]\[(2e-1)(e-1)\]
- Q2\[3k^2-4k+1\]\[(3k-1)(k-1)\]
- Q3\[5q^2-6q+1\]\[(5q-1)(q-1)\]
- Q4\[3u^2-5u+2\]\[(3u-2)(u-1)\]
- Q5\[6f^2-11f+3\]\[(3f-1)(2f-3)\]
- Q6\[4v^2-13v+3\]\[(4v-1)(v-3)\]
- Q7\[5g^2-16g+3\]\[(5g-1)(g-3)\]
- Q8\[4m^2-8m+3\]\[(2m-3)(2m-1)\]
- Q9\[3r^2-10r+8\]\[(r-2)(3r-4)\]
- Q10\[6x^2-13x+6\]\[(3x-2)(2x-3)\]

### Exercise 3: Factorising quadratics $a>1$, constant term $-$

**Instructions:**Factorise the following quadratic expressions- Q1\[10c^2+9c-9\]\[(5c-3)(2c+3)\]
- Q2\[2a^2-a-1\]\[(a-1)(2a+1)\]
- Q3\[2g^2+g-1\]\[(2g-1)(g+1)\]
- Q4\[4r^2+r-3\]\[(4r-3)(r+1)\]
- Q5\[3v^2-4v-4\]\[(v-2)(3v+2)\]
- Q6\[2x^2-x-6\]\[(x-2)(2x+3)\]
- Q7\[3h^2+5h-12\]\[(3h-4)(h+3)\]
- Q8\[3k^2-5k-12\]\[(k-3)(3k+4)\]
- Q9\[3b^2+b-10\]\[(3b-5)(b+2)\]
- Q10\[3x^2-x-10\]\[(x-2)(3x+5)\]

### Exercise 4: Mixed exercise

**Instructions:**Factorise the following quadratic expressions- Q1\[16n^2+12n-10\]\[2(2n-1)(4n+5)\]
- Q2\[9i^2-15i-36\]\[3(i-3)(3i+4)\]
- Q3\[20e^2+23e+6\]\[(5e+2)(4e+3)\]
- Q4\[10v^2-29v+10\]\[(2v-5)(5v-2)\]
- Q5\[36e^2-64\]\[4(3z-4)(3z+4)\]
- Q6\[4z^2+16z+15\]\[(2z+3)(2z+5)\]
- Q7\[14w^2+9w+1\]\[(7w+1)(2w+1)\]
- Q8\[5j^2-18j-8\]\[(j-4)(5j+2)\]
- Q9\[18h^2+15h+2\]\[(6h+1)(3h+2)\]
- Q10\[12r^2-14r-20\]\[2(r-2)(6r+5)\]