Factor (x + 1) to the fourth plus (x - 1) to the fourth, minus 82.

Step-by-step solution

1. 1

   Expand both fourth powers with the binomial theorem.

   $(x+1)^4=x^4+4x^3+6x^2+4x+1,\qquad (x-1)^4=x^4-4x^3+6x^2-4x+1$

2. 2

   Add them and watch the odd powers vanish.

   $(x+1)^4+(x-1)^4=2x^4+12x^2+2$

   This is not a coincidence: the second expression is the first with $x$ replaced by $-x$, and $f(x)+f(-x)$ always keeps only even-degree terms. Recognising this lets you skip the $x^3$ and $x$ bookkeeping entirely.

3. 3

   Subtract 82 and take out the common factor 2.

   $2x^4+12x^2+2-82=2x^4+12x^2-80=2\left(x^4+6x^2-40\right)$

4. 4

   Substitute $y=x^2$ to expose an ordinary quadratic.

   $y^2+6y-40=(y+10)(y-4)$

   since $10\cdot(-4)=-40$ and $10+(-4)=6$.

5. 5

   Back-substitute and finish factoring over the reals.

   $x^4+6x^2-40=(x^2+10)(x^2-4)=(x^2+10)(x-2)(x+2)$

   The factor $x^2+10$ has no real roots, so it stays intact; $x^2-4$ is a difference of squares and splits.

6. 6

   Collect and check numerically.

   $(x+1)^4+(x-1)^4-82=2(x-2)(x+2)(x^2+10)$

   At $x=1$: the original is $16+0-82=-66$, and $2(-1)(3)(11)=-66$ ✓. At $x=3$: $256+16-82=190$, and $2(1)(5)(19)=190$ ✓.