Expand the product of x² + 8x + 16 and x² - 4x + 4.

Step-by-step solution

1. 1

   Recognise each trinomial as a perfect square. Check the pattern $a^2\pm 2ab+b^2$:

   $x^2+8x+16=(x+4)^2,\qquad x^2-4x+4=(x-2)^2$

   The middle coefficients confirm it: $2\cdot 4=8$ and $2\cdot 2=4$. Nine separate term-by-term products are now avoidable.

2. 2

   Merge the two squares into one. Since $u^2v^2=(uv)^2$,

   $(x+4)^2(x-2)^2=\bigl[(x+4)(x-2)\bigr]^2$

   This is the whole trick: multiply the linear factors first, then square once.

3. 3

   Multiply the two binomials.

   $(x+4)(x-2)=x^2-2x+4x-8=x^2+2x-8$

   so the problem reduces to squaring the trinomial $x^2+2x-8$.

4. 4

   Square the trinomial with the three-term identity. Using $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$ with $a=x^2$, $b=2x$, $c=-8$:

   $a^2=x^4,\ b^2=4x^2,\ c^2=64,\ 2ab=4x^3,\ 2ac=-16x^2,\ 2bc=-32x$

5. 5

   Collect like terms. The two $x^2$ contributions combine:

   $x^4+4x^3+(4-16)x^2-32x+64=x^4+4x^3-12x^2-32x+64$

6. 6

   Check at $x=1$ and $x=-4$. At $x=1$ the original is $(1+8+16)(1-4+4)=25\cdot 1=25$, and the expansion gives $1+4-12-32+64=25$. At $x=-4$ the original is $0$ (since $(x+4)^2$ vanishes), and the expansion gives $256-256-192+128+64=0$. Both match.