Factorising large numbers

Question

Can I get some hint for this question, please?

I'm told to find the value of $p$ in the following equation

$$\dfrac{1}{(2p-1)^4} = \dfrac{1-2p}{32}$$

Here's my thought process ...

1. I can't possibly expand $(2p-1)^4$ as it will make things harder.
2. So I thought of changing the signs to $(1-2p)^4$
3. But ... $\dfrac{1}{(1-2p)^4}$ is not equal to $\dfrac{1}{(2p-1)^4}$ right?

Thanks for helping.

Answer 1

A simpler approach, and probably the intended solution: \begin{align} \frac{1}{(2p-1)^4}&=\frac{1-2p}{32}\\[2ex] (1-2p)(2p-1)^4&=32\\ (-1)(2p-1)(2p-1)^4&=32\\ (2p-1)^5&=-32\\ 2p-1&=-2\\ 2p&=-1\\ p&=-0.5 \end{align}

Answer 2

Why do you fear expansion ? The equation $\frac{1}{(2p-1)^4}=\frac{1-2p}{32}$ is equivalent to $$ (16p^4 - 48p^3 + 64p^2 - 52p + 31)(2p + 1)=0, $$ which has real solution $p=-\frac{1}{2}$, because the degree $4$ polynomial has no real roots.

Answer 3

$$\dfrac{1}{(2p-1)^4} = \dfrac{1-2p}{32}=\frac{1}{2^4(p-\frac 12)^4}=\frac{1-2p}{2^5}$$ Thus $$\frac{1}{(p-\frac 12)^4 }=\frac 12 -p\Rightarrow (p-\frac 12)^5=-1$$ Therefore $$p=-\frac12$$