1. ## factorise hayden

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

2. Originally Posted by johnsy123
$4x^2-4\sqrt{5x}+5$
Assuming a typo and you mean:

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

CB

3. Originally Posted by johnsy123
$4x^2-4\sqrt{5x}+5$
I assume this is actually

$4x^2 - 4\sqrt{5}x + 5$.

Remember that $(a + b)^2 = a^2 + 2ab + b^2$.

What you have posted is a perfect square, because

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

$= (2x - \sqrt{5})^2$