Question

factorise using g middle term
$2 \sqrt{2} {x}^{2} + 9x + 5 \sqrt{2 = }$
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Answer 1

Answer: $(x + \sqrt{2}) (2\sqrt{2} x + 5)$

Step-by-step explanation:

$2\sqrt{2} x^{2} + 9x + 5\sqrt{2}$

If there is an expression $ax^{2} +bx + c$ and we have to factorise it by middle term splitting, then we have to split the middle term in such a way that the

product of split terms = ac

sum of split terms = b

In this case,

a = 2√2 ; c = 5√2

2√2 × 5√2 = 20

So, we have to split the middle term, i.e., 9x in such a way that their product is 20x².

$2\sqrt{2} x^{2} + 9x + 5\sqrt{2}$

$= 2\sqrt{2} x^{2} + 4x +5x + 5\sqrt{2}$

4x + 5x is equal to 9x and 4x × 5x = 20x² So, we have met the conditions. Now we can take common and factorise the expression.

$2\sqrt{2} x^{2} + 9x + 5\sqrt{2}$

$= 2\sqrt{2} x^{2} + 4x +5x + 5\sqrt{2}$

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

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