How To Factorise Quadratic Expressions

Learn the basics of factorisation.

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In this post, we will show you how to factorise quadratics when \(a=1\) or \(a=-1\). Quadratics have the standard form:

$${ax^2+bx+c}$$

The procedure is as follows:

  1. Write your expression in standard form. Note the \(a\), \(b\) and \(c\) values.

  2. Make brackets such that \((\,x+f_1 )\,(\,x+f_2 )\,\), where \(f_1\) and \(f_2\) are factors of \(c\) (ignore the sign).

    The right factors are those that sum up to \(b\) (sign included).

  3. Look at the sign of \(c\).

    1. If \(c\) is positive, both factors take the sign of \(b\).
    2. If \(c\) is negative, the bigger factor takes the sign of \(b\)

Factorise the following:

  1. \(y=x^2-3x-54\)

    Factors of 54: \(1 \times 54\), \(2\times 27\), \(3 \times 18\), \(6 \times 9\). \(c\) is negative, so the bigger factor should take the sign of \(b\). The possible combination is: $${6 \times 9}$$ where \(6-9=-3\). Therefore:

    \begin{alignat} {1} y &= x^2-3x-54 \\ &= (\, x-9)\,(\,x+6 )\, \end{alignat}
  2. \(y=x^2-5x+6\)

    Using the procedure in (1.)

    \begin{alignat} {1} y &= x^2-5x+6 \\ &= (\, x-3)\,(\,x-2 )\, \end{alignat}
  3. \(y=-x^2-8x-15\)

    \begin{alignat} {1} y &= -x^2-8x-15 \\ &= -(\,x^2+8x+15 )\,\\ &= -(\, x+5)\,(\,x+3 )\, \end{alignat}

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