Question

factories it
$\sqrt{2x} + 7x + 5 \sqrt{2}$
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Answer 1

Given equation

$\sqrt{2}x^{2} +7x+5\sqrt{2} = 0$

Solving, we get,

$\sqrt{2}x^{2} +7x+5\sqrt{2} = 0$

$\implies \sqrt{2}x^{2} +2x+5x+5\sqrt{2} = 0$

$\implies \sqrt{2} x(x+\sqrt{2} )+5(x+\sqrt{2} ) = 0$

$\implies (\sqrt{2}x+5)(x+\sqrt{2}) = 0$

Now,

$\sqrt{2}x+5=0$

$\implies x = \dfrac{-5}{\sqrt{2} }$

$x+\sqrt{2} = 0\\\implies x = -\sqrt{2}$

$\boxed{\boxed{\bold{Therefore, \ the \ factorised \ expression \ is \ (\sqrt{2}x+5)(x+\sqrt{2}) }}}}}}}$

                                                                 

Answer 2

Given ,

The polynomial is √2x² + 7x + 5√2

By prime factorisation method ,

$\Rightarrow \sf \sqrt{2} {x}^{2} + 2x + 5x + 5 \sqrt{2} = 0 \\ \\ \Rightarrow \sf\sqrt{2} x(x + \sqrt{2}) + 5(x + \sqrt{2} ) = 0 \\ \\\Rightarrow \sf ( \sqrt{2} x + 5) (x + \sqrt{2}) = 0 \\ \\\Rightarrow \sf x = - \frac{5}{ \sqrt{2} } \: \: or \: \: x = - \sqrt{2}$

$\therefore \sf \underline{ \bold{The \: roots \: of \: polynomial \: are \: - \frac{ 5 }{ \sqrt{2} } \: and \: - \sqrt{2} }}$