Factorise x⁴ - 16 completely

Question

Factorise $x^4 - 16$ completely.

Solution — Step by Step

We have $x^4 - 16 = (x^2)^2 - 4^2$ .

This is a difference of two squares: $a^2 - b^2 = (a+b)(a-b)$ .

Here, $a = x^2$ and $b = 4$ .

x^4 - 16 = (x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)

Now look at each factor:

$(x^2 + 4)$ : This is a sum of a perfect square and a positive number. It cannot be factorised over real numbers (no real roots, since $x^2 + 4 \geq 4 > 0$ for all real $x$ ).

$(x^2 - 4)$ : This is again a difference of squares! $x^2 - 4 = x^2 - 2^2 = (x+2)(x-2)$ .

\boxed{x^4 - 16 = (x^2 + 4)(x + 2)(x - 2)}

This is the complete factorisation over real numbers — no factor can be simplified further.

Why This Works

The key is recognising that $x^4 - 16$ has a “nested” structure. First, it’s a difference of squares with respect to $x^2$ and 4. Then, the resulting factor $x^2 - 4$ is again a difference of squares with respect to $x$ and 2.

The formula $a^2 - b^2 = (a+b)(a-b)$ works because: $(a+b)(a-b) = a^2 - ab + ab - b^2 = a^2 - b^2$

Always keep applying the difference of squares formula until no factor can be broken down further. The sign of “sum vs difference” determines whether further factorisation is possible over real numbers.

Alternative Method

Direct substitution: Let $y = x^2$ . Then $x^4 - 16 = y^2 - 16 = (y-4)(y+4) = (x^2-4)(x^2+4)$ . Now substitute back and apply difference of squares to $(x^2-4)$ .

This substitution approach is cleaner for higher powers — it reduces unfamiliar expressions to familiar forms.

Common Mistake

The most common incomplete factorisation is stopping at $(x^2 + 4)(x^2 - 4)$ and not factorising $(x^2 - 4)$ further. The question asks for complete factorisation — every factorisable factor must be factorised. Check each factor: is it a difference of squares? Does it have common factors? Can it be factorised as a quadratic trinomial?

Over complex numbers, $(x^2 + 4) = (x + 2i)(x - 2i)$ where $i = \sqrt{-1}$ . But unless the problem specifies “factorise over complex numbers,” the real factorisation $(x^2 + 4)(x+2)(x-2)$ is the complete answer for Class 9–10 CBSE and JEE.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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