Factorise $2 x^{2} - 7 x - 39$

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Solution

We try to find two numbers $p$ and $q$ such that $p + q = - 7$ and $p \times q = 2 \times (- 39) = - 78$ . We can take $p = - 13$ and $q = 6$ as $- 13 + 6 = - 7$ and $(- 13) \times 6 = - 78$ .
Thus we write
$2 x^{2} - 7 x - 39 = 2 x^{2} - 13 x + 6 x - 39 = x (2 x - 13) + 3 (2 x - 13) = (x + 3) (2 x - 13)$

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