How to start this

Notice that we can factor the left-hand side of the equation as follows: $\begin{aligned} &\quad x^4 - 4x^2y + 3x^2y^2 + 5x^2 - 12y^3 + 15y^2 \\ &= x^2(x^2-4y) + 3x^2y^2 - 12y^3 + 5x^2 + 15y^2 \\ &= x^2(x^2-4y) + 3y^2(x^2-4y) + 5x^2 + 15y^2 \\ &= (x^2+3y^2)(x^2-4y) + 5(x^2+3y^2) \\ &= (x^2+3y^2)(x^2-4y+5). \end{aligned}$ Therefore, either $x^2+3y^2 = 0$ or $x^2-4y +5 = 0.$ However, since $x$ and $y$ are positive, $x^2+3y^2 > 0,$ so it must be the case that $x^2-4y +5 = 0.$ Therefore, $4y - x^2 = \boxed{5}.$

Note: I'd like to provide my thought process for the factoring above. To begin, I factored the first two terms just to see what happened, and bam! $x^2-4y$ shows up, which is the exact opposite of the quantity we want to find. We're making progress! Now I wanted to make more factors of $x^2-4y$ appear, and after a while then I found that $3x^2y^2-12y^3$ factors nicely, since the $-12$ is $-4$ times the $3.$ After that, I grouped the two terms together and finished off the last two terms, giving that nice factorization.

The given equation can be factorized as follows : $x^4-4x^2y+3x^2y^2+5x^2-12y^3+15y^2$

$=(x^4-4x^2y)+(3x^2y^2-12y^3)+(5x^2+15y^2)$

$=x^2(x^2-4y)+3y^2(x^2-4y)+5(x^2+3y^2)$

$=(x^2+3y^2)(x^2-4y)+5(x^2+3y^2)$

$=(x^2+3y^2)(x^2-4y+5)$

Since the given equation equals zero , we have

$(x^2+3y^2)(x^2-4y+5)=0$ which means either $x^2+3y^2=0$ or $x^2-4y+5=0$ . Considering the second condition , we get $x^2-4y+5=0 \Rightarrow 4y-x^2=\boxed{5}$

$0 =(x^4-4x^2y+5x^2)+(3x^2y^2-12y^3+15y^2)$

$0 =x^2(x^2-4y+5)+3y^2(x^2-4y+5)$

$0 =(x^2+3y^2)(x^2-4y+5)$

$x^2-4y+5=0$

$4y-x^2=\fbox{5}$

Note that

$\begin{aligned} x^4 - 4x^2y + 3x^2y^2 + 5x^2 - 12y^3 + 15y^2 &=\\ (x^2-4y+5)(x^2+3y^2) &= 0. \end{aligned}$

We know that $x$ and $y$ are strictly positive, so

$x^2+3y^2 > 0 \implies x^2 - 4y + 5 = 0,$

hence

$4y-x^2 = \boxed{5}.$

$x^{4}-4x^{2}y+3x^{2}y^{2}+5x^{2}-12y^{3}+15y^{2}=0$

$x^{4}+3x^{2}y^{2}+5x^{2}-4x^{2}y+15y^{2}-12y^{3}=0$

$x^{2}(x^{2}+3y^{2})+x^{2}(5-4y)+3y^{2}(5-4y)=0$

$x^{2}(x^{2}+3y^{2})+(5-4y)(x^{2}+3y^{2})=0$

$(x^{2}+3y^{2})(x^{2}-4y+5)=0$

$x^{2}-4y+5=0$

$4y-x^{2}=5$

The goal will be to factorize the expression into something with $4y - x^2$ as a factor.

Therefore, $x^4 - 4x^2 y + 3x^2 y^2 + 5x^2 - 12y^3 + 15y^2$ $= -x^2 (4y - x^2) - 3y^2 (4y - x^2) + 5(x^2 + 3y^2)$ $= -(4y - x^2)(x^2 + 3y^2) + 5(x^2 + 3y^2)$ $= (x^2 - 4y + 5)(x^2 + 3y^2) = 0$

Since $x$ and $y$ are positive real numbers, both $x^2$ and $3y^2$ will be positive. Thus, for the entire expression to be equals to $0$ , $x^2 - 4y + 5 = 0$ . Therefore, $x^2 - 4y = -5$ , and $4y - x^2 = 5$

$x^4 - 4x^2y + 3x^2y^2 + 5x^2 - 12y^3 + 15y^2 = 0$

can reduce to

$(x^2 + 3y^2)(4y - x^2 - 5)$ = 0

Therefore

So $4y - x^2 = 5$ .

Rearranging the equation, you can have $x^4 + 3x^2y^2- 4x^2y- 12y^3 + 5x^2 +15y^2= 0$ . Then, by factoring you can have $x^2(x^2 +3y^2) -4y(x^2 +3y^2 ) +5(x^2 + 3y^2)= 0$ . Hence, $(x^2 - 4y + 5)(x^2 +3y^2)= 0$ . $x^2 - 4y +5 = 0$ . $x^2 - 4y + = -5$ . $4y - x^2 = 5$

Factoring the given equation will result to:

$(x^{2} + 3y^{2})(x^{2} - 4y + 5) = 0$ .

Note that since $x$ and $y$ are both positive real numbers, the first factor does not make the product zero.

Therefore $x^2 - 4y + 5 =0$ . Rearranging the equation will give us $4y - x^{2} =5$

Therefore the answer is 5 .

The equation is equivalent to $(x^2+3y^2)(5-4y+x^2)=0$ . Since $x,y>0$ then $5-4y+x^2=0$ or $x^2-4y= \boxed{5}$ .

Let us modify the expression a bit:

$x^{4}-4x^{2}y+3x^{2}y^{2}+5x^{2}-12y^{3}+15y^{2}=0 \Rightarrow$

$\Rightarrow -x^{2}(4y-x^{2})-3y^{2}(4y-x^{2})+5(x^{2}+3y^{2})=0 \Rightarrow$

$\Rightarrow -(x^{2}+3y^{2})(4y-x^{2})+5(x^{2}+3y^{2})=0 \Rightarrow$

$\Rightarrow -(x^{2}+3y^{2})(4y-x^{2})=-5(x^{2}+3y^{2}) \Rightarrow$

$\Rightarrow 4y-x^{2}=\frac{-5(x^{2}+3y^{2})}{-(x^{2}+3y^{2})} \Rightarrow$

$\Rightarrow 4y-x^{2}=5$

From this, we can clearly see that $4y-x^{2}$ equals exactly $5$ , which is the correct answer.

Here we need to find the value of $4y-x^2$ , So let $4y-x^2=k \Rightarrow x^2=4y-k$ . Putting this in the given equation we get, $(4y-k)^2-4(4y-k)y+3(4y-k)^2y^2+5(4y-k)-12y^3+15y^2=0$ which on simplifying reduces to $k^2-4yk-3y^2k+20y-5k+15y^2=0$ $\Rightarrow 3(5-k)y^2+4(5-k)y+k(k-5)=0$ $\Rightarrow (5-k)(3y^2+4y-k)=0$ . Therefore we get $k=5$ . Hence 4y-x^2=5

x^2(x^2 - 4y) + 3y^2(x^2 - 4y) + 5(x^2 + 3y^2) = 0 (x^2 + 3y^2)(x^2 - 4y) + 5(x^2 + 3y^2) = 0 (x^2 - 4y + 5)(x^2 + 3y^2) = 0 x^2 - 4y + 5 = 0 4y - x^2 = 5

This problem can be solved with factorization:

$\begin{array}{l} {x^4} - 4{x^2}y + 3{x^2}{y^2} + 5{x^2} - 12{y^3} + 15{y^2} = 0\\ 3{x^2}{y^2} + 5{x^2} - 12{y^3} + 15{y^2} = {x^2}\left( {4y - {x^2}} \right)\\ 3{y^2}\left( {{x^2} - 4y} \right) + 5\left( {{x^2} + 3{y^2}} \right) = {x^2}\left( {4y - {x^2}} \right)\\ 3{y^2} + \frac{{5\left( {{x^2} + 3{y^2}} \right)}}{{\left( {{x^2} - 4y} \right)}} = - {x^2}\\ \frac{{5\left( {{x^2} + 3{y^2}} \right)}}{{\left( {{x^2} - 4y} \right)}} = - \left( {{x^2} + 3{y^2}} \right)\\ 5 = - \left( {{x^2} - 4y} \right)\\ 5 = 4y - {x^2} \end{array}$

x^4-4x^2y+3x^2y^2+5x^2-12y^3+15y^2=0,

x^4-4x^2y+5x^2 + 3x^2y^2-12y^3+15y^2=0,

x^2(x^2-4y+5)+3y^2(x^2-4y+5)=0,

(x^2+3y^2)(x^2-4y+5)=0,

x^2-4y+5=0,

or

4y-x^2=5

This expression can be factorised as (x^2+3y^2)(x^2-4y+5)=0 Therefore making x^2-4y+5=0 We get 4y-x^2=5

Notice that we can factor the left-hand side of the equation as follows: x4−4x2y+3x2y2+5x2−12y3+15y2=x2(x2−4y)+3x2y2−12y3+5x2+15y2=x2(x2−4y)+3y2(x2−4y)+5x2+15y2=(x2+3y2)(x2−4y)+5(x2+3y2)=(x2+3y2)(x2−4y+5). Therefore, either x2+3y2=0 or x2−4y+5=0. However, since x and y are positive, x2+3y2>0, so it must be the case that x2−4y+5=0. Therefore, 4y−x2=5.

We rewrite the equation as $(x^2-4y+5)(x^2+3y^2)=0$ and therefore $x^2-4y+5=0$ or $4y-x^2=5$

First, simplify or factor the equation..

( $x^{4}$ + 3 $x^{2}$ $y^{2}$ ) + ( 5 $x^{2}$ + 15 $y^{2}$ ) - ( 4 $x^{2}$ y - 12 $y^{3}$ ) = 0

then remove the common factor;

$x^{2}$ ( $x^{2}$ + 3 $y^{2}$ ) + 5( $x^{2}$ + 3 $y^{2}$ ) - 4y( $x^{2}$ + 3 $y^{2}$ = 0

and since the 3 has common factor, we can unite them using distributive property;

( $x^{2}$ - 4y + 5)( $x^{2}$ + 3 $y^{2}$ ) = 0

and the two factors can be equate to zero (but not both equal to zero)

we're looking for the value of ( 4y - $x^{2}$ );

            ( \\\( x^\{2\} \\\) - 4y + 5)
=>        4y - \\\( x^\{2\} \\\) = 5

By the Commutative Property, this equation also means $x^{4}-4x^{2}y+3x^{2}y^{2}-12y^{3}+5x^{2}+15y^{2} = 0$ . Simplifying, this equals $x^{2}(x^{2}-4y)+3y^{2}(x^{2}-4y)+5(x^{2}+3y^{2}) = (x^{2}+3y^{2})(x^{2}-4y)+5(x^{2}+3y^{2}) = 0$ . Therefore, $(x^{2}+3y^{2})(x^{2}-4y) = -5(x^{2}+3y^{2})$ , and $x^{2}-4y = -5$ . Finally, $4y-x^{2} = 5$ .