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Dear Uncle Colin,

I understand than $2x^3 - x^2 - 2x + 1$ factorises as $(2x-1)(x+1)(x-1)$, but I don’t know what steps to follow to get there. Any advice?

Presumably A Really Easy Ninja Trick Hacks Expression Simplification, Eh, Sensei?

Hi, PARENTHESES, and thanks for your message!

(On a point of order, I’m not sensei. The Mathematical Ninja is sensei. Maybe they’ll pop in to help later on.)

Unfortunately, factorisation is more of an art than a science. I can give you some tricks I might use, but they’re a bit ad-hoc because there isn’t a foolproof method for factorising cubics other than using some really horrendous formula like Cardan’s.

Inspection

“Inspection” is one of the words I hate as a mathematician; you might as well say “magic”. But here, there are some things you might spot.

There’s a 2 at the start of the first and third terms, and a 1 at the start of the second and last – that suggests rewriting it as $(2x^3 - 2x) - (x^2 - 1)$, then factorising the brackets separately to get $2x(x^2-1) - (x^2-1)$. Those, you can glom together as $(2x-1)(x^2-1)$, then use difference of two squares on the last bracket.

Alternatively, you might spot the alternating coefficients of 2 -1 -2 +1 and notice that the last pair is the same as the first pair, but with a flipped sign – then you would group as $(2x^3 - x^2) - (2x - 1)$, and continue from there: $x^2(2x -1) - (2x-1)$, or $(x^2-1)(2x-1)$, which again falls to the difference of two squares.

Factor theorem

Spotting that the constant term is 1 limits the possible “nice” factors – even with the leading coefficient of 2, the only values you would need to check are $x=\pm1$ and $x=\pm\frac{1}{2}$. It turns out that only one of these fails, and you can long-division your way out of trouble from there.

Make it numerical

AHARR!

Well, well, it’s the Mathematical Pirate. Haven’t seen you in a while, cap’n.

Arr. Shipwrecked. Had to tunnel me way out.

Obviously. Now, how would you approach this?

Let $x$ – which, of course, marks the spot – $=10$, and you’ll get an answer of 1881.

And that… helps?

It’s clearly a multiple of 9, which is $x-1$, and of 11, which is $x+1$.

… although a bit less clearly a multiple of 19, which is $2x-1$.

If you add 19 to it, you get 1900.

I suppose.

So you can write it as $(x-1)(x+1)(2x-1)$.

In the interests of balance, should we point out that you sometime get false factor this way – here, 3 is a factor, but $x-7$ isn’t. Goodness, you’re a bit quick with the cutlass, aren’t you?

Hope that helps!

- Uncle Colin