Question

2x^4-5x^2-3

Answer 1

Hi Casey,

I'm assuming you want to factor this since you didn't set it equal to zero. This is an example of "quadratic form," a concept that comes in very handy in certain situations like this one.

Basically, you are substituting a function of x into a standard quadratic equation. So you have \[au^2 + bu + c\]where u = f(x). Now u can be x^2, e^x, sin x... whatever. As long as the function is added to itself squared, it's quadratic form and you can factor/solve it like you would any other quadratic.

In your case, u = x^2, so you can think of you problem as \[2(x^2)^2-5(x^2)-3\] Now, since you know that 2u^2-5u-3 factors into (2u+1)(u-3), you can immediately write
\[(2x^2+1)(x^2-3)\]I'll let you take it from here: use the zero-factor theorem to solve each of these for x, or factor them further by setting them equal to zero, solving, and using the roots: (x-r1)(x-r2).

Hope this helps!