I'm struggling with factoring these two questions.

36x^4 - 3x^3 - 18x^2
9x^2 + 3xy - 2y^2

36x^4 - 3x^3 - 18x^2

3x^2 (12x^2-x-6)
3x^2 (4x-3)(3x+2)

9x^2 + 3xy - 2y^2
(9x+)(x-) won't work; the 9 is too big

3*2 - 3*1 = 3, so
(3x+2)(3x-1)

To factor the given expressions, we'll be using a technique called factoring by grouping for the first expression and factoring using the difference of squares for the second expression.

Let's begin with the first expression:
36x^4 - 3x^3 - 18x^2

Step 1: Look for a common factor among all the terms. In this case, the common factor is 3x^2.
3x^2(12x^2 - x - 6)

Step 2: Now we need to factor the quadratic expression (12x^2 - x - 6). We'll look for two numbers whose product is -72 (product of the leading coefficient and the constant term) and whose sum is -1 (coefficient of x term). The numbers are -9 and 8.
3x^2(12x^2 - 9x + 8x - 6)

Step 3: Group the terms:
3x^2 [(12x^2 - 9x) + (8x - 6)]

Step 4: Factor out the common terms from each group:
3x^2 [3x(4x - 3) + 2(4x - 3)]

Step 5: Now, notice that we have (4x - 3) common in both groups. So we can factor it out:
3x^2(4x - 3)(3x + 2)

Therefore, the factored form of 36x^4 - 3x^3 - 18x^2 is 3x^2(4x - 3)(3x + 2).

Moving on to the second expression:
9x^2 + 3xy - 2y^2

This expression is a quadratic expression, so we'll use the technique called factoring using the difference of squares.

Step 1: Identify the terms that can be written as perfect squares. In this case, 9x^2 and -2y^2 are perfect squares.

Step 2: Apply the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b).
In this case, a = 3x and b = y.
9x^2 - 2y^2 can be written as (3x)^2 - (y)^2, which is in the form of a^2 - b^2.

Step 3: Apply the difference of squares formula to factor the expression:
(3x + y)(3x - y)

Therefore, the factored form of 9x^2 + 3xy - 2y^2 is (3x + y)(3x - y).