Factorise $10 x^{2}-x-24$ by splitting the middle term.
Factorise $10 x^{2}-x-24$ by splitting the middle term.
If we can find two numbers $p$ and $q$ such that $p+q=-1$ and
$p q=(10)(-24)=-240,$ then we can get the factors.
$p=15$ and $q=-16$ satisfy $p+q=1$ and $p q=-240$
So, $10 x^{2}-x-24$
$=10 x^{2}+15 x-16 x-24$
$=5 x(2 x+3)-8(2 x+3)$
$=(2 x+3)(5 x-8)$
$10 \times(-24)=-240$
$(-240)=15 \times(-16)$
$15+(-16)=-1$
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