# Wrong, But Useful: Episode 58 | Colin

In this episode of Wrong, But Useful, we are joined by freelance mathematician @becky_k_warren, formerly of NRICH
Becky likes sharing maths with people who “don’t like maths” and the #beingmathematical twitter chat
Number of the podcast: 157, which is the middle of a sexy prime triplet. Colin goes all Rees-Mogg for a moment.
The Aperiodical Math-off is reaching its final stages. We were rooting for @ajk_44. Dave suggests a televised version. Colin dislikes the tournament format.
@cmonmattthink has More Questions Than Answers
@cuttheknotmath (Alexander Bologmony in real life) has died….

https://ift.tt/2uGYLuO

# Why the factor and remainder theorem work | Colin

So there I was, merrily teaching the factor and remainder theorems, and my student asked me one of my favourite questions: “I accept that the method works, but why does it?”

(I like that kind of question because it makes me think on my feet in class, and that makes me feel alive!)
Let’s get everyone up to speed, shall we?

The factor theorem says that $(x-a)$ is a factor of a polynomial $p(x)$ if and only if $p(a)=0$. That means, if a question says “show that $x-7$ is a factor of $8x^3 -58x^2 – x + 105$”, all you need to do is work out what happens to that expression when you stick a 7…

https://ift.tt/2L1kpF6

Dear Uncle Colin,

I had to find the $n$th term of a quadratic sequence (1, 6, 17, 34, 57). I remember my teacher saying something about a table, but I couldn’t figure it out. Can you help?

Struggles Expressing Quadratics Using Educator’s Notation – Concrete Explanation?

Hi, SEQUENCE, and thank you for your message!

(Before beginning, I should point you at Mark’s article on this topic from some time ago.)

I have used table-based methods in the past, but – just like you say – they don’t seem to stick in the memory. Instead, I much prefer an algebra-based solution. It’s a little…

https://ift.tt/2N6iwnn

# Two coins, one fair, one biased | Colin

When the redoubtable @cuttheknotmath (Alexander Bogomolny) poses the following question:

Two Coins: One Fair, one Biased https://t.co/Rz2zR3LRDj #FigureThat #math #probability pic.twitter.com/HHhnyGjhkq

— Alexander Bogomolny (@CutTheKnotMath) March 5, 2018

… you know there must be Something Up. Surely (the naive reader thinks) the one with two heads out of three is the one with a probability of two heads in three? But equally surely (thinks the reader who knows Alexander doesn’t set trivial problems), it can’t be that simple.
So let’s work it out.

We have two coins, label them A…

https://ift.tt/2m59Nqc

# Ask Uncle Colin: A Coin Toss Conundrum | Colin

Dear Uncle Colin,

Please can you settle an argument? I say, if you toss a coin three times, the probability of getting all heads is one in four, because the only possibilities are HHH, HHT, HTT and TTT. My friend says it’s one in eight, being $\frac{1}{2}\times \frac{1}{2} \times \frac{1}{2}$. Who’s right, and why?

Tiffing Over Some Sums

Hi, TOSS, and thanks for your message!

I’m afraid it’s your friend who’s correct. Their interpretation is spot on: the probability of getting a head is a half, and the probability of tossing three in a row is $\br{\frac{1}{2}}^3$.

The more…

https://ift.tt/2MM2TBm

# A calculator puzzle | Colin

“Your calculator has broken, leaving you with only the buttons for $\sin$, $\cos$, $\tan$ and their inverses, the equals button and the 0 that starts on the screen. Show that you can still produce any positive rational number.”

When this showed up on Reddit, I knew I was in for a) a rough ride and b) a bit of a treat. (The poster mentioned that it was an International Mathematical Olympiad problem, and those are often the worst kind of fun.)

In this post, I want to talk not just about the solution, but the nebulous process that took me there.

Below the line be spoilers.
My eventual…

https://ift.tt/2KmtrMR

# Ask Uncle Colin: An Unclear Inequality | Colin

Dear Uncle Colin,

I solved $(x+1)(x-2)(x+3)>0$ by saying there were three possibilities, $x+1>0$, $x-2>0$ or $x+3>0$. The middle one gives $x>2$ and that’s the strictest, so that was my answer – but apparently it’s wrong. Why is that?

– Logical Expressions Seem Silly

Hi, LESS, and thank you for your message!

I have one piece of advice for you, one:
Sketch. Sketch, sketch, sketch.

I though it was such an important piece of advice that it was worth repeating four times.

The most critical thing you can do when working with an inequality is to sketch it. See where it…