Factor Theorem - Deep Thinking

Using the table mode on a calculator allows you to find the roots of a polynomial quickly, and hence factorise it with minimal effort. There's an opportunity here to explore some aspects of polynomials and develop a deeper understanding at A-Level.

Solution

Substituting any positive value for \(x\) will result in the sum of positive values. The only way to achieve \(p(x)=0\) is to have at least one negative term, which comes from cubing a negative value or multiplying a negative value by 20.

Solution

Utilise the fact from the previous question; all roots must be negative because of the positive signs. Also, the linear factors must each have constant terms which multiply to 9, too. That is, if \(p(x)=(x-a)(x-b)(x-c)\) then \(-abc=9\). We either have -1, -1, -9 or -1, -3, -3 as the three roots. In both cases, there is a repeated root.

Solution

Dhriti is using the leading coefficient of 2 to anticipate a factor in the form \((2x+m)\). The assosciated root will be \(x=-\frac{m}{2}\) so using a step of 0.5 will reveal when \(p(x)=0\).

Similarly, for a leading coefficient of 3, Dhriti could use a step of \(\frac{1}{3}\) to spot when \(p(x)=0\).

 

Solution

Hint 1: One bracket must start with \(2x\) and the others are just \(x\)

Hint 3: The signs are all positive so the roots are all negative. The linear brackets will therefore be \((2x+a)(x+b)(x+c)\) where \(a, b, c >0\)

Hint 2: The values \(a\), \(b\) and \(c\) must multiply to 2. They will be 1, 1 and 2 in some order. \(a\neq 2\) because \(2x+2\) has a common factor of 2, meaning all coefficients of \(p(x)\) would be even. \(b\) and \(c\) can be assigned 1 and 2 without loss of generality.

So we get \((2x+1)(x+1)(x+2)\)
 

Inequality Puzzle - [Futoshiki]

I find that students assosciate > with the words "greater than" and assosciate that with "bigger", as if it goes through a "terminology phase" in their head. Two steps means it's twice as likely to go wrong!

I wanted to get rid of the words and let students get used to how the inequality signs interact with the values directly. The words can come later on.

So, here is a worksheet with 8 inequality puzzles. Place the numbers 1 to n in the grid such that each number appears exactly once in each row and column. There are some inequality signs that also have to be obeyed. I've found that Key Stage 3 classes spend 15-30 minutes on it.

Here's a teaser version.

Since creating this and using it a few times, I've learned that this type of puzzle is called Futoshiki, translating to "More or Less". If you want more, there are plenty out there. I think this one is pretty good and have used this on the board as a time filler in lessons. Just a heads up that some require an awful lot of logic and deduction which may be too much for some students, whereas the ones in my worksheet were aimed at getting used to inequalities with a gradual increase in logic.

HCF with a Calculator

Calculators are being given more functionality with each upgrade. Prime factorisation has been around for quite a while, but can calculators find the HCF of two numbers for you?

Yes, but not directly.

Enter two numbers as a fraction and the calculator will simplify it. It's just divided by the HCF. In this case, 36 and 80 become 9 and 20, which are a quarter of the size, so the highest common factor of 36 and 80 is 4.

I wouldn't say this is particularly useful for (AQA) Higher tier students, as basic HCF questions don't come up in calculator papers. As for foundation, AQA 83003F 2022 asked for the HCF of 12 and 18, and AQA 83002F 2020 asked for the HCF of 75 and 105. For some students, this trick might save them a couple of marks.

I'm posting this idea mainly to be used as a discussion point in lessons, to get students thinking deeper about the Maths. 

What about the LCM?

You can use the idea that \(LCM(a,b) = \frac{ab}{HCF(a,b)}\) 

In this case \(\frac{36\times80}{4}=720\)

I don't see that as a shortcut and probably involves a deeper understanding of HCF and LCM than the listing method or comparing them in index form. Again, it's a discussion point to deepen understanding.

A final discussion...

Is it possible to use the calculator in a similar way to find the HCF of 3 numbers? 

Circle in the Valley

A circle rests on the vertex of the curve \(y=x^2\). What radii can the circle have if it only touches the curve at one point? Here's a Desmos graph set up to play around with...

Solution

Force the circle \(x^2+(y-R)^2=R^2\) and the curve \(y=x^2\) to intersect at exactly one point.

Then replacing \(x^2\) with \(y\), we get \(y+(y-R)^2=R^2 \implies y+y^2-2Ry+R^2=R^2\\ \implies y^2+(1-2R)y=0\\ \implies y(y+1-2R)=0\).

\(y=0\) is a solution. The circle and curve are set up to intersect at this point.

The second bracket being zero gives \(y=-(1-2R)\) and we know \(y>0\) based on the initial setup so \(1-2R<0 \implies R>\frac{1}{2}\) for a solution.

 If \(y>0\) we do not want any further solutions, so we must have \(0<R\leq \frac{1}{2}\).

 ---

Let's take this further. What about a circle resting in the valley of a cosine curve? Here's another Desmos graph with a circle on the curve \(y=1-cos(x)\) for convenience...

Solution

The previous method now involves solving equations containing both \(x\) and \(cos(x)\). Maybe reattempt the previous question with a different method prior to tackling this one.

Jumping back to the \(y=x^2\) question. We know the gradient of the circle and the curve are both 0 at the origin. Other than this point, provided the gradient of the circle is always greater than the gradient of the curve, the circle remains above the curve. Let's calculate the gradients.

The circle is \(f(x)=R-\sqrt{R^2-x^2}\) so \(f^{\prime}(x)= \frac{x}{\sqrt{R^2-x^2}}\) .
The curve has gradient \(2x\).

Force the gradient of the circle to be greater than the gradient of the curve.

\(\frac{x}{\sqrt{R^2-x^2}}<2x\) 

We know that \(x=0\) is where the gradients are equal, so ignore this point and say \(x\neq0\). Divide through by \(2x\) and multiply through by \(\sqrt{R^2-x^2}\) to get 

\(\frac{1}{2}>\sqrt{R^2-x^2} \implies R^2<\frac{1}{4}+x^2\)

This needs to hold true for \(0<x<R\). Very close to \(x=0\) gives \(R^2<\frac{1}{4} \implies R<\frac{1}{2}\).

--- 

Jumping back to the \(y=1-cos(x)\) curve and applying this calculus method.

The circle is \(f(x)=R-\sqrt{R^2-x^2}\) so \(f^{\prime}(x)=\frac{x}{\sqrt{R^2-x^2}}\).
The curve is \(g(x)=1-cos(x)\) so \(g^{\prime}(x)=sin(x)\)

If \(f^{\prime}(x)>g^{\prime}(x)\) then \(\frac{x}{\sqrt{R^2-x^2}}>sin(x) \implies x>sin(x) \sqrt{R^2-x^2}\).

As \(x\) approaches \(R\), this is clearly true as \(\sqrt{R^2-x^2}\) tends to 0.
As \(x\) becomes very small, say \(\delta x\) then this becomes \(\delta x>sin(\delta x) \sqrt{R^2-{\delta x}^2}\) and using small angle approximations, we have \(\delta x>\delta x \sqrt{R^2-{\delta x}^2} \implies \sqrt{R^2-{\delta x}^2} <1 \implies R<1\).

 

Subtraction Calculation Investigation

In the middle of a lesson on subtraction, I stumbled across a nice looking calculation. Later, I investigated variations of it and spotted some nice patterns.

Below, you can the first slide of a PowerPoint I've put together, which can be found here.

There are five solutions, each of which has \(R=3\): \(51-15=36, 62-26=36, 73-37=36, 84-48=36, 95-59=36\)

Let's change it slightly. If the visible digit was 5 instead of 6, how would that change the answer?

There are four solutions, each of which has \(R=4\): \(61-16=45, 72-27=45, 83-38=45, 94-49=45\) 

What about if the digit were 4, 3 and so on? 

The pattern continues with \(R\) and the visible digit summing to 9, and the number of solutions being 1 less than the visible digit.

What about the other direction? What about the digit being 7, 8 and so on?

The pattern continues up to 8, but the visible digit being 9 requires \(R=0\) which isn't allowed. Similarly, the digit being 0 requires \(W=S\) resulting in 00.

I've already explained the pattern, but the PowerPoint has a slide prompting students to try identify the pattern. There's also a nice discussion about extrapolation here (which is why I started at 6 and worked outwards!)

Can you prove that \(R\) and the visible digit always sum to 9?

This is a nice little proof for students with an understanding of how to convert numbers like \(AB\) into \(10A+B\).

If \(W=S\) then the result is 0, so reject that idea. If \(W<S\) then the result is negative, so reject this too. We now know that \(W>S\).

Looking at the ones column, \(W>S\) so one of the tens needs exchanging to ten ones. In old money, we need to borrow from the \(W\).

The calculations become:

Ones column: \(10+S-W=X \implies 10-X=W-S\)
Tens column: \((W-1)-S=R \implies W-S=R+1\)

Equating the \(W-S\) gives \(10-X=R+1\implies R=9-X\) as needed. 

Trial and Improvement

Trialling \(x=7\) gives \(252\) which is too small and \(x=8\) gives \(320\) which is too big. The positive solution must be one of these two, but which one?

From my experience, students tend to look at \(252\) and \(320\) and use those to determine which one is closer, but they should actually trial \(x=7.5\) giving \(285\) which is too small and therefore conclude that \(x=8\). For years, I struggled to explain why they need to do this, because this shortcut seems to work for an overwhelming proportion of questions, so I set out to find something that broke that habit.

Well, the question above does just that. \(252\) and \(320\) are both equidistant from the \(286\) so it provides no additional information. Students have to rely on testing \(x=7.5\)

I did produce a series of questions that did this, but it seems that every question I created, the midpoint trial value was always too small and so it always rounded up. I don't want to create new misconceptions, but did want to share this one example as a wake-up call for those not going to one extra level of accuracy.

If I can generate some questions that round down, expect a worksheet.

My Baby and I are Similar

My baby and I have the same right footprint (I'll leave that to the imagination) and someone commented that "he's probably going end up being my height when he's older - is he on track?"

Naturally, I was curious.

Assuming similarity between my baby and me, he should be 78cm tall. He is actually 68cm tall, give or take. This could spark discussion around model limitations. For example, my baby doesn't stand with straight legs yet and there are other genetic factors at play.