Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\\-(t-2)(t+2)^3+(t+2)^4\right]=242t$$ Let $A=(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-(t-2)(t+2)^3+(t+2)^4.$
Then $$A=(t-2)^4-(t-2)^2(t^2-4+t^2+4t+4)-(t+2)^3(2-t+t+2)\\=(t-2)^4-2t(t+2)(t-2)^2-4(t+2)^3.$$
 A: We have
$$0=(x-1)^5+(x+3)^5-242(x+1)=2(x^2 + 2x + 42)(x + 2)(x + 1)x
$$
by applying the rational root theorem to the polynomial equation
$$
2x^5 + 10x^4 + 100x^3 + 260x^2 + 168x=0,
$$
which yields the linear factors $x$, $x+1$ and $x+2$.
A: That's a helpful start.
Notice $(t-2)^5+(t+2)^5$ is an odd function:
$$ ((-t)-2)^5+((-t)+2)^5 = -(t+2)^5-(t-2)^5 = -\left((t-2)^5+(t+2)^5\right) $$
Therefore $A = \frac{(t-2)^{5}+(t+2)^{5}}{2t}$ is an even function, so when it's multiplied out and collected to a basic polynomial form, that form must be $A = Bt^4+Ct^2+D$. This is encouragement that it won't be that bad, plus it means we'll be able to finish the problem by considering $A-121$ as a quadratic in $t^2$.
I get $$ A = t^4+40t^2+80 $$ so the equation is
$$ 2t(t^4+40t^2+80) = 242 t$$
$$ 2t(t^4+40t^2-41) = 0 $$
$$ t(t^2-1)(t^2+41) = 0 $$
The solutions (three real and one complex pair) are $t \in \{-1, 0, 1, i\sqrt{41}, -i\sqrt{41}\}$, $x \in \{-2,-1,0,-1+i\sqrt{41},-1-i\sqrt{41}\}$.
A: The “brute-force” way is to use the Binomial Theorem:
$$(x+a)^5 = x^5 + 5ax^4 + 10a^2x^3 + 10a^3x^2 + 5a^4x + a^5$$
Which gives us:
$$(x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1) + (x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243) = 242x + 242$$
$$2x^5 + 10x^4 + 100x^3 + 260x^2 + 410x + 242 = 242x + 242$$
$$2x^5 + 10x^4 + 100x^3 + 260x^2 + 168x = 0$$
$$2x(x^4 + 5x^3 + 50x^2 + 130x + 84) = 0$$
Clearly, $x=0$ is a solution.  By the rational root theorem, roots of the quartic factor must be a factor of 84 (i.e., in $\pm\{ 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84\}$).  Descartes' rule of signs rules out all positive solutions.  Trial and error gives $x = -1$ and $x = -2$ as roots.
Dividing the quartic by the factors $(x+1)(x+2)$ gives the quadratic equation $x^2 + 2x + 42$, with the roots $-1 \pm i \sqrt{41}$.
Putting this all together, $x \in \{ 0, -1, -2, -1 + i\sqrt{41}, -1 - i\sqrt{41} \}$.
A: You could also solve it in a much simpler way, like this:
$$
\begin{align*}
(x - 1)^{5} + (x + 3)^{5} &= 242 \cdot (x + 1) \quad\mid\quad -(242 \cdot (x + 1))\\
(x - 1)^{5} + (x + 3)^{5} - 242 \cdot (x + 1) &= 0\\
2 \cdot x^{5} + 10 \cdot x^{4} + 100 \cdot x^{3} + 260 \cdot x^{2} + 168 \cdot x &= 0 \quad\mid\quad \div 2\\
x^{5} + 5 \cdot x^{4} + 50 \cdot x^{3} + 130 \cdot x^{2} + 84 \cdot x &= 0\\
x \cdot (x^{4} + 5 \cdot x^{3} + 50 \cdot x^{2} + 130 \cdot x + 84) &= 0 \quad\mid\quad \text{factor into four terms}\\
x \cdot (x + 1) \cdot (x + 2) \cdot (x^{2} + 2 \cdot x + 42) &= 0 \quad\mid\quad \text{use the zero product theorem}\\
\\
x &= 0 \tag{1.}\\
x + 1 &= 0 \quad\mid\quad -1 \tag{2.}\\
x + 2 &= 0 \quad\mid\quad -2 \tag{3.}\\
x^{2} + 2 \cdot x + 42 &= 0 \quad\mid\quad -41 \tag{4.}\\
\\
x &= 0 \tag{1.}\\
x &= -1 \tag{2.}\\
x &= -2 \tag{3.}\\
x^{2} + 2 \cdot x + 1 &= -41 \tag{4.}\\
\\
x &= 0 \tag{1.}\\
x &= -1 \tag{2.}\\
x &= -2 \tag{3.}\\
(x + 1)^{2} &= -41 \quad\mid\quad \sqrt{~~~} \tag{4.}\\
x + 1 &= \pm\sqrt{41} \cdot \mathrm{i} \quad\mid\quad -1\\
\\
x &= 0 \tag{1.}\\
x &= -1 \tag{2.}\\
x &= -2 \tag{3.}\\
x &= -1 \pm \sqrt{41} \cdot \mathrm{i} \tag{4.}\\
\end{align*}
$$
