I'm having troubles with these problems from a packet I got:
1. 1/x^2-9
2. 2(x-1)^2-(x-1)-1
3. y^2(a-b)-(a-b)
4. c^2+c+1/4
5. (x+1)^2-3(x-1)+2
6. x^4-9x^2+20
7. y^2-1/6y-1/6
8. n^4+2n^2-3
Please help. Thanks.
Sure, I'd be happy to help you with these problems. Let's go through each problem step by step.
1. 1/x^2-9
To simplify this expression, we need to factor the denominator, which is the expression x^2 - 9. This is a difference of squares, so we can rewrite it as (x - 3)(x + 3). Now the expression becomes 1/((x - 3)(x + 3)).
2. 2(x-1)^2-(x-1)-1
First, we can simplify the expression inside the parentheses by expanding (x - 1)^2, which gives us (x - 1)(x - 1) = x^2 - 2x + 1. Now the expression becomes 2(x^2 - 2x + 1) - (x - 1) - 1. Next, distribute the 2 to the terms inside the parentheses, giving us 2x^2 - 4x + 2 - x + 1 - 1. Finally, combine like terms to simplify the expression further.
3. y^2(a-b)-(a-b)
In this expression, we have common factors in both terms: (a - b). We can factor it out to simplify the expression. After factoring out (a - b), the expression becomes (y^2 - 1)(a - b).
4. c^2+c+1/4
This expression doesn't appear to be factorable, so we can simply leave it as is.
5. (x+1)^2-3(x-1)+2
We can start by expanding (x + 1)^2, which gives us x^2 + 2x + 1. Now the expression becomes x^2 + 2x + 1 - 3(x - 1) + 2. Next, distribute the -3 to the terms inside the parentheses, giving us x^2 + 2x + 1 - 3x + 3 + 2. Finally, combine like terms to simplify the expression further.
6. x^4-9x^2+20
This expression doesn't appear to be factorable, so we can simply leave it as is.
7. y^2-1/6y-1/6
To simplify this expression, we need to have a common denominator for the fractions in the expression. We can multiply both the numerator and denominator of 1/6 by 6 to get a common denominator of 6. Now the expression becomes y^2 - 6/6y - 6/6. Next, combine the fractions and simplify the expression.
8. n^4+2n^2-3
This expression doesn't appear to be factorable, so we can simply leave it as is.
I hope this helps! Let me know if you have any further questions.