In this post, we will show you how to factorise quadratics when \(\normalsize a=1\) or \(\normalsize a=-1\). Quadratics have the standard form:
$${\normalsize ax^2+bx+c}$$Factorising quadratics of this form is quite easy. Once you grasp the method, it's easy to solve any other quadratics. The procedure is as follows:
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Write your expression in standard form. Note the \(\normalsize a\), \(\normalsize b\) and \(\normalsize c\) values.
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Make brackets such that \((\,\normalsize x+f_1 )\,(\,\normalsize x+f_2 )\,\), where \(\normalsize f_1\) and \(\normalsize f_2\) are factors of \(\normalsize c\) (ignore the sign).
The right factors are those that sum up to \(\normalsize b\) (sign included).
Look at the sign of \(\normalsize c\).
- If \(\normalsize c\) is positive, both factors take the sign of \(\normalsize b\).
- If \(\normalsize c\) is negative, the bigger factor takes the sign of \(\normalsize b\)
Factorise the following:
\(\normalsize y=x^2-3x-54\)
Factors of 54: \(\normalsize 1 \times 54\), \(\normalsize 2\times 27\), \(\normalsize 3 \times 18\), \(\normalsize 6 \times 9\). \(\normalsize c\) is negative, so the bigger factor should take the sign of \(\normalsize b\). The possible combination is: $${\normalsize 6 \times 9}$$ where \(\normalsize 6-9=-3\). Therefore:
$$ \normalsize \begin{alignat}{1} y &= x^2-3x-54 \\ &= (\, x-9)\,(\,x+6 )\, \end{alignat} $$\(\normalsize y=x^2-5x+6\)
Using the procedure in (1.)
$$ \normalsize \begin{alignat}{1} y &= x^2-5x+6 \\ &= (\, x-3)\,(\,x-2 )\, \end{alignat} $$\(\normalsize y=-x^2-8x-15\)
$$ \normalsize \begin{alignat}{1} y &= -x^2-8x-15 \\ &= -(\,x^2+8x+15 )\,\\ &= -(\, x+5)\,(\,x+3 )\, \end{alignat} $$