The polynomial expressions $$ 4 x^{2}+8 x y-5 y^{2} x+24 x^{2}-6 y^{2} $$ sture which of the following factors

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We begin with the polynomial written as a sum of two groups:
$$
(4x^2+8xy-5y^2) + (24x^2-6y^2).
$$

Step 1. Factor the first group. Notice that
$$
4x^2+8xy-5y^2
$$
is a quadratic in $$x$$ (with coefficients involving $$y$$). We look for two numbers $$m$$ and $$n$$ such that
$$
2x\cdot 2x=4x^2,\quad m+n=8y,\quad \text{and} \quad m\cdot n=-5y^2.
$$
A good choice is $$m=2x\cdot 5y$$ and $$n=2x\cdot (-y)$$ so that:
$$
(2x- y)(2x+5y)=4x^2+ (2x\cdot 5y- y\cdot 2x) -5y^2 =4x^2+8xy-5y^2.
$$

Step 2. Factor the second group. Factor out the common factor in
$$
24x^2-6y^2:
$$
we have
$$
24x^2-6y^2=6(4x^2-y^2).
$$
Notice $$4x^2-y^2$$ is a difference of two squares:
$$
4x^2-y^2=(2x)^2-y^2=(2x-y)(2x+y).
$$
Thus,
$$
24x^2-6y^2=6(2x-y)(2x+y).
$$

Step 3. Write the whole expression as the sum:
$$
(2x-y)(2x+5y)+6(2x-y)(2x+y).
$$
Now, factor out the common factor $$2x-y$$:
$$
(2x-y)\Big[(2x+5y)+6(2x+y)\Big].
$$

Step 4. Simplify the expression in the brackets:
$$
(2x+5y)+6(2x+y)=2x+5y+12x+6y=(2x+12x)+(5y+6y)=14x+11y.
$$

Thus, the original polynomial factors as:
$$
(2x-y)(14x+11y).
$$

The factorization shows that one of the factors is $$\boxed{2x-y}$$ and the other factor is $$\boxed{14x+11y}$$.
The polynomial factors into $$ (2x - y)(14x + 11y) $$.

The polynomial expressions sture
which of the following factors

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The polynomial expressions sture which of the following factors