Rewrite the expression x^2 + 8xy + 16y^2 - 4z in a simpler form.

Step-by-step solution

1. 1

   Test the first three terms for a perfect square. A trinomial $A^2 + 2AB + B^2$ needs its outer terms to be squares and its middle term to be twice their product. Here $x^2 = (x)^2$ and $16y^2 = (4y)^2$, and

   $2 \cdot x \cdot 4y = 8xy$

   which matches the middle term exactly.

2. 2

   Write the square.

   $x^2 + 8xy + 16y^2 = (x + 4y)^2$

3. 3

   Reassemble with the remaining term.

   $x^2 + 8xy + 16y^2 - 4z = (x + 4y)^2 - 4z$

4. 4

   Explain why this is as far as it goes. The form $(x+4y)^2 - 4z$ is a difference of squares only if $4z$ is itself a square, i.e. $z = w^2$ for some expression $w$; then it would factor as $\left(x + 4y - 2w\right)\left(x + 4y + 2w\right)$. With $z$ an unrestricted independent variable, no further factorisation over the polynomial ring exists.

5. 5

   Check numerically. At $(x, y, z) = (3, -1, 2)$: the original gives $9 - 24 + 16 - 8 = -7$, and $(3 - 4)^2 - 8 = 1 - 8 = -7$. At $(2, 1, 0)$: $4 + 16 + 16 - 0 = 36$ and $(2+4)^2 = 36$. Both match.