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 \( \normalsize f(x) = 2x^2-9x+9\).
Methods:
-
Draw up a table of coefficients.
Use the factor theorem to find \(f(x) = 0\). In this example, \(x=3\) will work, i.e., \(\normalsize f(3)=2(3)^2-9(3)+9=0\).
3 2 -9 9 0 6 -9 2 -3 0 Explanation
- The first row is the coefficients of the quadratic, except the first element which is the root(vakue that makes the expression equal zero.
- The second and third rows of the first column should be left empty.
- Put a zero on the row 2, column 2.
- To get the value of row3, column 2, add the values of the row 1 and 2 in the same column. In this case, coulmn 2.
- To get the value of row 2, column 3, multiply row3, column 2 with the value of row1,column 1.
- Repeat the steps to fill the remaining parts.
The final answer is the factor multiplied by the first two values of the third row, i.e.,
$$ \normalsize f(x)=(x-3)(2x-3)$$ -
Multiply the factor with some linear expression \( (mx+p)\).
$$ \normalsize \begin{alignat}{1} f(x) &= 2x^2-9x+9\\ f(3) &= 0 \end{alignat} $$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,
$$\normalsize f(x)=(x-3)(2x-3)$$ -
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.
As with every technique, practice makes perfect. Practice, you will master these.