### Instruction: Completely factor the following expression: \[(15x^3+80x-5)-(-4x^3+4x-5).\].### Response: I want to factor this expression, so I first need to simplify it by removing the parentheses and combining like terms.
To remove the parentheses, I need to distribute the negative sign to the second group of terms, which means changing the sign of each term inside.
I get \[15x^3+80x-5+4x^3-4x+5.\]
Now I can combine the like terms, which are the ones with the same power of x.
I have two terms with x^3, two terms with x, and two terms with no x.
When I combine them, I get \[19x^3+76x.\]
This is much simpler than the original expression, but I still want to factor it completely.
To factor something completely means to write it as a product of prime factors, or factors that cannot be factored further.
One way to look for factors is to see if there is a common factor in all the terms.
In this case, I notice that both terms have a factor of x, and also a factor of 19.
So I can factor out 19x from both terms, and write \[19x^3+76x=19x(x^2+4).\]
This is partially factored, but I'm not done yet.
I need to check if the remaining factors, 19x and x^2+4, can be factored further.
19x cannot be factored further, because 19 is a prime number and x is a variable.
x^2+4 cannot be factored further using real numbers, because there are no two real numbers that multiply to 4 and add to 0.
However, if I allow complex numbers, which involve the imaginary unit i=\sqrt{-1}, then I can factor x^2+4 as (x+2i)(x-2i).
This is because (x+2i)(x-2i) expands to x^2-4ix+4ix-4i^2, and -4i^2 is the same as 4, since i^2=-1.
So, using complex numbers, I can factor the expression completely as \[19x^3+76x=19x(x+2i)(x-2i).\]
This is the final answer, unless the problem only wants real factors, in which case I would stop at \[19x^3+76x=19x(x^2+4).\]