(a) For the function $f(x)=x^{4}+361$, factor $f$ into the product of two irreducible quadratics. (b) Find the zeros of $\mathrm{f}$ by finding the zeros of each irreducible quadratic. (a) $f(x)=$ (Factor completely. Type an exact answer, using radicals as needed.)

Step 1 ：Rewrite the function \(f(x)=x^{4}+361\) as \(f(x)=x^{4}-(-19)^2\).

Step 2 ：Recognize this as a difference of squares, which can be factored as \((x^2-19)(x^2+19)\).

Step 3 ：Note that these are not irreducible quadratics over the real numbers, as they can be factored further.

Step 4 ：The first factor can be factored as \((x-\sqrt{19})(x+\sqrt{19})\), and the second factor can be factored as \((x-i\sqrt{19})(x+i\sqrt{19})\), where \(i\) is the imaginary unit.

Step 5 ：However, these are not quadratics, so the original factors \((x^2-19)\) and \((x^2+19)\) are the irreducible quadratics over the real numbers.

Step 6 ：\(\boxed{f(x)=(x^2-19)(x^2+19)}\) is the final answer.