# Factorise Quadratics When $$a \neq \pm 1$$

Factorising quadratics when $$a \neq \pm 1$$.

In the last blog post, we showed you how to factorise quadratics when $$a = \pm 1$$. Click here to be directed to that post.

Today, we will show you how to factorise quadratics when $$a \neq \pm 1$$. We will give you three methods and you will choose whichever one suits you best and then, go ace that grade 10, 11 or 12 exam.

We will use an example of $$f(x) = 2x^2-9x+9$$.

## Methods:

1. Draw up a table of coefficients.

Use the factor theorem to find $$f(x) = 0$$. In this example, $$x=3$$ will work, i.e., $$f(3)=2(3)^2-9(3)+9=0$$.

3 2 -9 9
0 6 -9
2 -3 0

Explanation

• In the first column, put a zero in the middle row.
• To get the third row elements, add the first two rows.
• To get the remaining row two elements, multiply the preceding row three elements with the factor, in this case, 3. Continue in this manner.

The final answer is the factor times the first two values of the third row,i.e.,

$${f(x) = (\,x-3 )\, (\,2x-3)\,}$$
2. Multiply the factor with some linear expression $$(\, mx + p )\,$$.

\begin{alignat} {1} f(x) &= 2x^2-9x+9\\ f(3) &= 0 \end{alignat}

Now we have:

$${f(x) = (\,x-3 )\, (\,mx+p)\,}$$

Now let, $$m$$ be the $$a$$ value and let $$p$$ multiplied by -3 be $$c$$. Therefore, $$m=2$$ and $$-3p=9 \Rightarrow p=-3$$.Finally,

$${f(x) = (\,x-3 )\, (\,2x-3)\,}$$
3. Break down the middle term of the quadratic and perform normal factorisation.

• Find the product of $$a \times c$$ (signs ignored). $$2 \times 9 = 18$$
• Find factors of the above product. $$1 \times 18, 2 \times 9, 3 \times 6$$.
• Note the conditions of $$c$$ stated in the previous lesson. The factors we will use are 3 and 6.
\begin{alignat} {1} f(x) &= 2x^2-9x+9\\ f(x) &= 2x^2-3x-6x+9\\ f(x) &= x(\,2x-3)\,-3(\,2x-3)\, \\ f(x) &=(\,2x-3)(\,x-3)\, \end{alignat}

As with every technique, practice makes perfect. Practice, you will master these.