Factorize $: 2 x^{2} - 7 x + 5 = 0$

$(2 x - 7) (x - 1) = 0$

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$(x - 2) (x - \frac{7}{2}) = 0$

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$(x - 2) (x - \frac{7}{5}) = 0$

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$(2 x - 5) (x - 1) = 0$

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Solution

The correct option is D
$(2 x - 5) (x - 1) = 0$

We need to find two numbers whose product is $(5 \times 2) = 10$ and whose sum is $- 7$

Let us first split the middle term $- 7 x$ as $- 5 x - 2 x$
$[∵ (- 5 x) \times (- 2 x) = 10 x^{2} = (2 x^{2}) \times 5]$
So, $2 x^{2} - 7 x + 5 = 2 x^{2} - 5 x - 2 x + 5 = x (2 x - 5) - 1 (2 x - 5) = (2 x - 5) (x - 1)$

Now, $2 x^{2} - 7 x + 5 = 0$ can be rewritten as $(2 x - 5) (x - 1) = 0$ .

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