state how many linear factors each polynomial has: My school is literally screwing me over please frickin help

To find the number of linear factors of the polynomial $f\left(x\right)=x^{16}+72x^6+x$, we can use the fact that every polynomial with complex coefficients can be factored into linear and quadratic factors.

First, let's check if there are any quadratic factors by computing the discriminant of the quadratic equation obtained by setting $x^{16}+72x^6+x$ equal to zero. The discriminant is $b^2-4ac$, where a = 1, b = 72, and c = 1. Plugging these values into the formula, we get:

$b^2-4ac={72}^2-4\left(1\right)\left(1\right)=5184-4=5180$

Since the discriminant is positive, the quadratic equation has two complex roots, which means that f(x) cannot be factored into quadratic factors.

Next, we can look for linear factors by checking for roots of the polynomial. Unfortunately, there is no easy way to find the roots of a polynomial of degree 16. However, we can use the fact that f(x) has complex coefficients to our advantage.

Recall that complex roots of a polynomial with real coefficients always come in conjugate pairs. That is, if a + bi is a root of the polynomial, then so is its conjugate a - bi. Since f(x) has complex coefficients, any non-real roots must come in conjugate pairs. Therefore, if we can find one non-real root of f(x), we automatically obtain eight linear factors, corresponding to the conjugates of the root.

One way to find a non-real root is to use the intermediate value theorem. Note that $f\left(0\right)=0+0+0=0,\text{and}f\left(1\right)=1+72+1=74$. Since f(x) is a continuous function, there must be some value of x between 0 and 1 for which f(x) is equal to a complex number. Therefore, f(x) has at least one non-real root, which gives us eight linear factors.

In summary, the polynomial $f\left(x\right)=x^{16}+72x^6+x$ cannot be factored into quadratic factors and has at least eight linear factors (corresponding to the conjugates of a non-real root).