Factor:

2(2x^2-x)^2-3(2x^2-x)-9

i also need help with this question factor:

(x+y)^2-x-y

help mee with the two questions, i don't get it and i have a test tomorrow!

someone please help!!!!

To make it easier to see,

let 2x^2 - x = a
so you have
2a^2 - 3a - 9
= (a-3)(2a+3)
put the original back in
= (2x^2 - x - 3)(4x^2 - 2x + 3)
= (x + 1)(2x - 3)(4x^2 - 2x + 3)

(x+y)^2-x-y

= (x+y)^2 - (x+y)
= (x+y)(x + y - 1)

To factor the expression 2(2x^2-x)^2 - 3(2x^2-x) - 9, we can follow these steps:

Step 1: Factor out the common factor. Notice that the entire expression has a common factor of (2x^2 - x). We can rewrite the expression by factoring out this common factor:
Common factor = (2x^2 - x)
2(2x^2 - x)^2 - 3(2x^2 - x) - 9 = (2x^2 - x)(2(2x^2 - x) - 3) - 9

Step 2: Simplify the expression inside the parentheses. Multiply the common factor (2x^2 - x) by the remaining factors:
= (2x^2 - x)(4x^2 - 2x - 3) - 9

Step 3: Factor the expression (4x^2 - 2x - 3). To do this, we will look for two numbers, let's say "a" and "b", that multiply to give -12 (the product of the coefficient of x^2 and the constant term) and add up to -2 (the coefficient of x).
The numbers that fit these criteria are -4 and 3 because -4 * 3 = -12 and -4 + 3 = -1.
So, we can rewrite the expression:
= (2x^2 - x)(4x^2 - 4x + 3x - 3) - 9

Step 4: Group the terms. We can now group the terms within the second set of parentheses:
= (2x^2 - x)[(4x^2 - 4x) + (3x - 3)] - 9

Step 5: Factor out the common factors within each group. Notice that the terms (4x^2 - 4x) and (3x - 3) both have a common factor of 4x and 3, respectively. We can factor them out:
= (2x^2 - x)(4x(x - 1) + 3(x - 1)) - 9

Step 6: Simplify the expression inside the parentheses. Now, we have a common factor of (x - 1) between the two terms within the second set of parentheses. We can factor it out:
= (2x^2 - x)(4x(x - 1) + 3(x - 1)) - 9
= (2x^2 - x)(4x + 3)(x - 1) - 9

So, the completely factored form of the given expression is (2x^2 - x)(4x + 3)(x - 1) - 9.