How would you do practise test 3, section 3, question 16? I plugged in 1 and happened to get the answer. But is there an algebraic way to do it?

Thanks!

Sure. Algebraically, you can do it in a few steps:

    \begin{align*}x^3(x^2-5)&=-4x\\x^5-5x^3&=-4x\\x^5-5x^3+4x&=0\\x(x^4-5x^2+4)&=0\\x(x^2-4)(x^2-1)&=0\end{align*}

You can actually go further by expanding the differences of two squares if you want (below), but because the question tells you that x> 0, you know from the above that it can either be 1 or 2.

    \begin{align*}x(x^2-4)(x^2-1)&=0\\x(x+2)(x-2)(x+1)(x-1)&=0\end{align*}

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