How to factorise it? Can you explain to me how to factorise it? and I should remove a common factor before factorising the trinomials.
This question...
$$
4x^2 + 12xy +9y^2
$$
 A: Consider $$4x^2+12xy+9y^2$$
$$=4x^2+6xy+6xy+9y^2$$
$$=(4x^2+6xy)+(6xy+9y^2)$$
What common factor can you take out from each bracket?
A: $$4x^2 + 12xy + 9y^2$$
$$= (2x)^2 + 2(2x)(3y) + (3y^2)$$
Write formula $$(a + b)^2 = a^2 + 2ab + b^2$$Then evaluate $a = 2x$ and $b = 3y$. Are you done with it?
A: the discriminant of $a x^2 +bxy + c y^2 $ is
$$  b^2 - 4ac  $$
same as in the Quadratic formula. With integer abc and any common integer factor pulled out, especially any $-1,$ so that $a \geq 0 \; \; : \; \;$ if the discriminant is a positive square, the thing factors nicely. If, in addition, the discriminant is exactly $0,$  the thing itself is a square. Since 4 and 9 are squares, the choices are
$$ (2x-3y)^2 $$
$$ (2x+3y)^2 $$
A: Denote this function as $f(x,y)=4x^2+12xy+9y^2$ and let it equal $0$. You can treat either $x$ or $y$ as constant, and reduce this as a function of either $x$ or $y$. For example, I treat it as a function of $x$ and consider the discriminant of the function:
$$\Delta=144y^2-4\times 4\times 9y^2=0$$
which implies this function has only one root of order $2$, which is $$x=-\frac{3}{2}y$$
Hence $$f(x,y)=\left(2x+3y\right)^2$$
