Find all roots of x^4 - 16 = 0, both the real roots and the complex ones.

Step-by-step solution

1. 1

   Spot the difference of two squares. Both terms are perfect squares once you look at them the right way:

   $x^{4}=(x^{2})^{2},\qquad 16=4^{2},$

   so the identity $a^{2}-b^{2}=(a-b)(a+b)$ applies with $a=x^{2}$ and $b=4$:

   $x^{4}-16=(x^{2}-4)(x^{2}+4).$

   Factoring beats taking fourth roots directly because it exposes every root, including the complex ones.

2. 2

   Factor again where possible. The first bracket is another difference of squares:

   $x^{2}-4=(x-2)(x+2),$

   while $x^{2}+4$ is a sum of squares and does not factor over the real numbers. The full real factorisation is therefore

   $x^{4}-16=(x-2)(x+2)(x^{2}+4).$

3. 3

   Use the zero-product property on the linear factors. A product is zero exactly when one factor is zero:

   $x-2=0\Rightarrow x=2,\qquad x+2=0\Rightarrow x=-2.$

   These are the only two real roots — which is what you should expect, since $x^{4}=16$ has exactly two real solutions, $\pm\sqrt[4]{16}=\pm 2$.

4. 4

   Handle the remaining quadratic over the complex numbers. Setting $x^{2}+4=0$ gives $x^{2}=-4$, which has no real solution but two imaginary ones:

   $x=\pm 2i.$

   So over $\mathbb{C}$ the complete factorisation is $(x-2)(x+2)(x-2i)(x+2i)$, and the degree-$4$ polynomial has exactly $4$ roots, as the fundamental theorem of algebra requires.

5. 5

   Verify each root. $2^{4}=16$ ✓ and $(-2)^{4}=16$ ✓. For the imaginary pair, $(2i)^{4}=2^{4}i^{4}=16\cdot 1=16$ ✓ and likewise $(-2i)^{4}=16$ ✓. All four roots lie on a circle of radius $2$ in the complex plane, evenly spaced at $90^{\circ}$ — the standard picture for the fourth roots of a positive real number.