Determine 2 values of k so that trinomials can be factored over integers 36m^2 + 8m + k and 18x^2 - 42y + k.
To determine the values of k for which the given trinomials can be factored over integers, we need to find the factors of the leading coefficient and the constant term.
1. For the trinomial 36m^2 + 8m + k:
The factors of 36 are: ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36.
The factors of k can vary depending on its value, so we will keep it as a variable for now.
2. For the trinomial 18x^2 - 42y + k:
The factors of 18 are: ±1, ±2, ±3, ±6, ±9, ±18.
The factors of k can vary depending on its value, so we will keep it as a variable for now.
By inspecting the factors of the leading coefficient and the constant term, we need to find two values of k where both the trinomials can be factored over integers.
Let's consider some examples:
For the first trinomial, let's try k = 1:
36m^2 + 8m + 1 = (6m + 1)(6m + 1) = (6m + 1)^2
For the second trinomial, let's try k = 9:
18x^2 - 42y + 9 = 3(6x^2 - 14y) + 9 = 3(2x)^2 - 3(2y)^2 + 9 = (2x - 3(2y))^2
So, for the values k = 1 and k = 9, both trinomials can be factored over integers.
To determine the values of k such that the trinomials can be factored over integers, we need to find the factors of the quadratic equations and look for combinations that result in integer solutions.
Let's start with the first trinomial: 36m^2 + 8m + k.
To factor this trinomial, we look for two integers, let's call them a and b, such that the sum of their product and the product of a^2 and b^2 is equal to the quadratic coefficients.
In this case, the coefficient of m^2 is 36, the coefficient of m is 8, and there is no constant term. So, we need to find two integers whose product is 36 and whose sum is 8.
By examining the factors of 36, we can see that 6 and 6 are the only integers whose product is 36. However, their sum is 12, not 8.
To compensate, we can try different combinations of factors to see if any of them satisfy the conditions. Let's try factors of 36 and see if their sum is equal to 8:
1 * 36 = 36 (Sum = 37)
2 * 18 = 36 (Sum = 20)
3 * 12 = 36 (Sum = 15)
4 * 9 = 36 (Sum = 13)
6 * 6 = 36 (Sum = 12)
None of these satisfy the condition of having a sum equal to 8.
So, for the trinomial 36m^2 + 8m + k, there are no integer values of k such that it can be factored over integers.
Now let's move on to the second trinomial: 18x^2 - 42y + k.
Similarly, we need to find two integers whose sum is -42 (the coefficient of y) and whose product is equal to the product of the coefficients of x^2 (18) and y (-42).
Again, by examining the factors, we can see that there are no integers that satisfy these conditions.
For the trinomial 18x^2 - 42y + k, there are no integer values of k such that it can be factored over integers.
In conclusion, there are no values of k for either trinomial that allow for factoring over integers.
Not quite sure what you're getting at, but
36m^2 + 8m - 28 = (9m-7)(4m+4)
In fact, given any n,
(9m - (9n+7))(4m + (4n+4)) = 36m^2 + 8m + (36n^2 + 64n + 28)
Assuming that the second one is 42x, not 42y,
(9x-3)(2x-4) = 18x^2 - 42x + 12
I'm sure you can find many more
There's probably some number-theoretic principles that will enable you to find a general formula, but I'd have to study up a bit.