Factorize 72×2-10x-42.

# Factorize $7\sqrt{2}{x}^{2}-10x-4\sqrt{2}$.

1. A
$\left(7\sqrt{2}x-4\right)\left(x-\sqrt{2}\right)$
2. B
$\left(7\sqrt{2}x+4\right)\left(x-\sqrt{2}\right)$
3. C
$\left(7\sqrt{2}x+4\right)\left(x+\sqrt{2}\right)$
4. D
$\left(7\sqrt{2}x+4\right)\left(x-\sqrt{2}\right)$

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### Solution:

Given,$7\sqrt{2}{x}^{2}-10x-4\sqrt{2}$.
Factorising, by splitting the middle term whose sum is -10 and product is -56.$=7\sqrt{2}{x}^{2}-14x+4x-4\sqrt{2}$
= $7\sqrt{2}x\left(x-\sqrt{2}\right)+4\left(x-\sqrt{2}\right)$ $=\left(7\sqrt{2}x+4\right)\left(x-\sqrt{2}\right)$
Thus, factors are $\left(7\sqrt{2}x+4\right)\left(x-\sqrt{2}\right)$.
Therefore, option 2 is  correct.

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