SOLUTION: Find the constant k so that 2x^2 + 5xy - 8y^2 + 7x + 25y + k can be expressed as the product of two linear factors of the form ax + by + c.

Question 1209763: Find the constant k so that
2x^2 + 5xy - 8y^2 + 7x + 25y + k
can be expressed as the product of two linear factors of the form ax + by + c.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let the given expression be
$$2x^2 + 5xy - 8y^2 + 7x + 25y + k$$
We want to express this as a product of two linear factors of the form $(ax + by + c)(dx + ey + f)$.
First, factor the quadratic terms:
$$2x^2 + 5xy - 8y^2 = (2x - y)(x + 8y)$$
So, we expect the linear factors to be of the form:
$$(2x - y + c_1)(x + 8y + c_2)$$
Expanding this, we get:
$$2x^2 + 16xy + 2c_2x - xy - 8y^2 - c_2y + c_1x + 8c_1y + c_1c_2$$
$$2x^2 + 15xy - 8y^2 + (2c_2 + c_1)x + (8c_1 - c_2)y + c_1c_2$$
We are given that the expression is
$$2x^2 + 5xy - 8y^2 + 7x + 25y + k$$
Comparing the coefficients of $xy$, we see that $15xy \neq 5xy$, so the factorization $(2x - y)(x + 8y)$ is incorrect.
Let's try a different factorization of $2x^2 + 5xy - 8y^2$.
We need to find two numbers that multiply to $2(-8) = -16$ and add to $5$. These numbers are $8$ and $-2$, so we can rewrite the middle term as $8xy - 3xy$.
$$2x^2 + 8xy - 3xy - 12y^2 = 2x(x + 4y) - 3y(x + 4y) = (2x - 3y)(x + 4y)$$
So we expect the linear factors to be of the form:
$$(2x - 3y + c_1)(x + 4y + c_2)$$
Expanding this, we get:
$$2x^2 + 8xy + 2c_2x - 3xy - 12y^2 - 3c_2y + c_1x + 4c_1y + c_1c_2$$
$$2x^2 + 5xy - 12y^2 + (2c_2 + c_1)x + (4c_1 - 3c_2)y + c_1c_2$$
Comparing coefficients:
$$2c_2 + c_1 = 7$$
$$4c_1 - 3c_2 = 25$$
$$c_1c_2 = k$$
Multiply the first equation by 4:
$$8c_2 + 4c_1 = 28$$
Subtract the second equation from this:
$$8c_2 + 4c_1 - (4c_1 - 3c_2) = 28 - 25$$
$$11c_2 = 3$$
$$c_2 = \frac{3}{11}$$
Substitute $c_2 = \frac{3}{11}$ into $2c_2 + c_1 = 7$:
$$2(\frac{3}{11}) + c_1 = 7$$
$$\frac{6}{11} + c_1 = 7$$
$$c_1 = 7 - \frac{6}{11} = \frac{77 - 6}{11} = \frac{71}{11}$$
Now, find $k = c_1c_2$:
$$k = (\frac{71}{11})(\frac{3}{11}) = \frac{213}{121}$$
Thus, the constant $k$ is $\frac{213}{121}$.
Final Answer: The final answer is $\boxed{213/121}$

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