If a factor has $x^2$, is it considered a linear factor?

Question

So I have been dealing with this question:

Factor $f(x)=2x^4-2x^3+4x^2-8x-16$ completely into linear factors.

After attempting the question, my calculations result in: $f(x)=2(x+1)(x-2)(x^2+4)$.

As far as I know of what is meant by factor into linear factors, the factors should have linear graphs if they were to be graphed separately. In this case, $x^2+4$ does not have a linear graph but rather the graph of a parabola. Is my understanding of linear factorization incorrect or is it possible that $x^2+4$ can be factored using complex numbers?

Thank you ahead of time!

Answer 1

I think what is intended here is to use complex numbers so that you really get linear factors. You have to write $x^{2}+4$ as $(x-2i)(x+2i)$.

Answer 2

You are entirely right that $x^2 + 4$ isn't considered a linear factor, and it can be factored in the complex numbers (any polynomial can be factored in the complex numbers; that's called the fundamental theorem of algebra). You factor it just like you would any other quadratic: Find its roots: $$ \frac{0\pm\sqrt{0 - 4\cdot 1\cdot 4}}{2\cdot 1} = \pm 2i $$ and insert that into $(x-x_1)(x-x_2)$ to get $x^2 - 4 = (x-2i)(x+2i)$.

Also, you forgot the leading coefficient of $2$. So the correct factorisation would be $$ f(x) = 2(x+1)(x-2)(x-2i)(x+2i) $$

Answer 3

The intention is that each factor in your decomposition would have $x^1$ (as opposed to $x^2$) so as to be linear; thus, yes, the intention is to decompose using complex numbers where necessary.

In turn, we should replace $x^2+4$ with $(x+2i)(x-2i)$. You can see this from the difference-of-two-squares formula by noting $x^2 + 4 = x^2 - (-4)$.

Answer 4

Well, linear factors are of the form $x-a$, where $a$ is a number. So your factorization has a quadratic term $x^2+4$. If you work over the rational or real numbers, it cannot be factored further. However, each polynomial over the rationals or reals can be factored in linear factors over the complex numbers, since the field of complex numbers is algebraically closed. Here you can factor $x^2+4 = (x-2i)(x+2i)$, since the latter equals $x^2-(2i)^2$ and $(2i)^2 = 4i^2=-4$.

The complete factorization is then $2(x+1)(x-2)(x-2i)(x+2i)$.