Factor the expression (x + 1)^4 + (x + 3)^4 - 82 completely.

Step-by-step solution

1. 1

   Substitute at the midpoint so the expression becomes symmetric. The two bases $x+1$ and $x+3$ are centred on $x+2$, so setting $y=x+2$ makes them $y-1$ and $y+1$. Expanding directly in $x$ would produce a messy quartic; this choice is what makes the odd terms cancel:

   $(y-1)^4+(y+1)^4-82$

2. 2

   Expand and watch the odd powers cancel.

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

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

   so the expression becomes $2y^4+12y^2+2-82=2y^4+12y^2-80=2\left(y^4+6y^2-40\right)$.

3. 3

   Treat the bracket as a quadratic in $t=y^2$. With $t=y^2$,

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

   since $10\cdot(-4)=-40$ and $10+(-4)=6$. Back in $y$: $\left(y^2+10\right)\left(y^2-4\right)$.

4. 4

   Factor the difference of squares and stop at the irreducible piece. $y^2-4=(y-2)(y+2)$, while $y^2+10$ has no real roots and stays intact:

   $2\left(y^2+10\right)(y-2)(y+2)$

5. 5

   Substitute $y=x+2$ back. $y-2=x$, $y+2=x+4$, and $y^2+10=(x+2)^2+10=x^2+4x+14$:

   $2x(x+4)\left(x^2+4x+14\right)$

6. 6

   Verify by evaluating both forms at several integers. Checking $x=-8,\dots,8$, the original $(x+1)^4+(x+3)^4-82$ and $2x(x+4)(x^2+4x+14)$ agree at every one of those $17$ points — far more than the $5$ needed to pin down a quartic, so the factorization is exact.