# Factorising large numbers

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.

• With $1-2p/32$ do you mean $1-p/16$ or $(1-2p)/32$ ? – Giovanni Resta Feb 22 '16 at 15:34
• $\dfrac{1}{(1-2p)^4}$ is equal to $\dfrac{1}{(2p-1)^4}$, because $(-x)^4=x^4$ for all $x$. But that doesn't help, unless you have written the right-hand side incorrectly. Perhaps it should be $(1-2p)/32$? – TonyK Feb 22 '16 at 15:35

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.
• I think the equation is supposed to be $\frac{1}{(2p-1)^4}=\frac{1-2p}{32}$ – Logophobic Feb 22 '16 at 15:34
$$\dfrac{1}{(2p-1)^4} = \dfrac{1-2p}{32}=\frac{1}{2^4(p-\frac 12)^4}=\frac{1-2p}{2^5}$$ Thus $$\frac{1}{(p-\frac 12)^4 }=\frac 12 -p\Rightarrow (p-\frac 12)^5=-1$$ Therefore $$p=-\frac12$$
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