is problem number 1 correct

Problem #1
factor completely. x^4 + x^3 -12

MY answer: (x^2-3)(x^2+4) or is this still factorable.

Problem #2

i need someone to help me set it up for me to solve.

If the sides of a square are decreased by 3cm, the area is decreased by 81 cm^2.What were the dimensions of the original square?

No, you do not have it factored. How did the x cubed term get there? If the x^3 term should be x^2, then it is correct.

(x-3)^2=x^2-81
solve for x.

sorry typo it should read as

x^4+x^2-12

To factor the expression x^4 + x^2 - 12, we can start by looking for common factors among the terms. In this case, we don't have any common factors like numbers or variables.

To proceed, we can try factoring the expression by grouping. Let's group the terms together:

(x^4 + x^2) - 12

Now, let's try to factor out a common factor from each group separately.

In the first group, we can factor out x^2, giving us:

x^2(x^2 + 1) - 12

Now, we have a difference of squares in the first group, which can be factored further. The difference of squares formula states that:

a^2 - b^2 = (a + b)(a - b)

In our case, a = x and b = 1, so we have:

x^2(x + 1)(x - 1) - 12

Simplifying further, we have the fully factored expression:

(x + 1)(x - 1)(x^2 - 12)

So, the correct factorization of x^4 + x^2 - 12 is (x + 1)(x - 1)(x^2 - 12).