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Factorize $8 x^{2}-10 x-3$
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We need to find two numbers such that their product is $-24$ and their sum is $-10$. Let's find the factors of $-24$.

Our desire numbers are $-12$ and 2 .

We will now re-write our trinomial: $8 x^{2}-12 x+2 x-3$

$=\left[8 x^{2}-12 x\right]+[2 x-3]$

$=4 x(2 x-3)+1(2 x-3)$

$=(4 x+1)(2 x-3)$

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