Category Archives: Arithmetic

July 1, 2025 · 10:35 am

Synthetic Division (for factorising and/or solving polynomials)

I use synthetic division to factorise polynomials with a degree greater than 2. For example, $f(x)=x^3+2x^2-5x-6$

It works best with monic polynomials but can be adapted to non-monic ones (see example below).

The only problem is that you need to find a root to start.

Try the factors of $-6$ i.e. $(-1, 1, 2, -2, 3, -3, 6, -6)$

$f(-1)=(-1)^3+2(-1)^2-5(-1)-6=0$

Hence, $x=-1$ is a root and $(x+1)$ is a factor of the polynomial.

Set up as follows

Bring the first number down

Multiply by the root and place under the second co-efficient

Add down

Repeat the process

The numbers at the bottom $(1, 1, -6)$ are the coefficients of the polynomial factor.

We now know $x^3+2x^2-5x-6=(x+1)(x^2+x-6)$ .

We can factorise the quadratic in the usual way.

$x^2+x-6=(x+3)(x-2)$

Hence $x^3+2x^2-5x-6=(x+1)(x+3)(x-2)$ .

Let’s try a non-monic example

Factorise $6x^4+39x^3+91x^2+89x+30$

I know $-2$ is a root. Otherwise I would try the factors of 30.

Use synthetic division

Because this was non-monic we need to divide our new co-efficients $(6, 27, 37, 15)$ by 6 (the co-efficient of the $x^4$ term)

$x^3+\frac{9}{2}x^2+\frac{37}{6}+\frac{5}{2}$

We now need to go again. I know that $\frac{-3}{2}$ is a root and $(2x+3)$ is a factor.

Our quadratic factor is $x^2+3x+5/3$ , which is $3x^2+9x+5$ .

The quadratic factor doesn’t have integer factors so,

$6x^4+39x^3+91x^2+89x+30=(x+2)(2x+3)(3x^2+9x+5)$

I think this is much quicker than polynomial long division.

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Filed under Algebra, Factorising, Factorising, Fractions, Polynomials, Quadratic, Simplifying fractions, Solving Equations

Tagged as non monic synthetic division, Synthetic division

June 19, 2025 · 7:18 am

Converting recurring (non-terminating) decimals to fractions

The easiest approach is to jump right in with some examples.

Example 1

Convert $0.\overline{5}$ to a fraction.

Let $x=0.\overline{5}$

(1) $\begin{equation*}x=0.\overline{5}\end{equation*}$

(2) $\begin{equation*}10x=5.\overline{5}\end{equation*}$

Subtract equation $1$ from equation $2$

$\begin{equation*}9x=5\end{equation}$

Hence $x=\frac{5}{9}$ so $0.\overline{5}=\frac{5}{9}$

Example 2

Convert $0.\overline{12}$ to a fraction.

Let $x=0.\overline{12}$

(3) $\begin{equation*}x=0.\overline{12}\end{equation*}$

(4) $\begin{equation*}100x=12.\overline{12}\end{equation*}$

Subtract equation $3$ from equation $4$ .

$\begin{equation*}99x=12\end{equation}$

$\begin{equation*}x=\frac{12}{99}=\frac{4}{33}\end{equation}$

Example 3

Convert $0.1\overline{23}$ to a fraction

Let $x=0.1\overline{23}$

(5) $\begin{equation*}x=0.1\overline{23}\end{equation*}$

(6) $\begin{equation*}10x=1.\overline{23}\end{equation*}$

(7) $\begin{equation*}1000x=123.\overline{23}\end{equation*}$

Subtract equation $6$ from equation $7$

$\begin{equation*}990x=122\end{equation}$

$\begin{equation*}x=\frac{122}{990}=\frac{61}{495}\end{equation}$

Our aim is to manipulate the recurring decimal to create two numbers each which have only the repeated digits after the decimal point.

One more example.

Example 4

Convert $3.4\overline{56}$ to a fraction

Let $x=3.4\overline{56}$

If I multiply by 10, then I will have $34.\overline{56}$ – only repeated digits after the decimal point.

If I multiply by 1000, then I will have $3456.\overline{56}$ – only repeated digits after the decimal point.

So I get,

$\begin{equation*}990x=3422\end{equation}$

$\begin{equation*}x=\frac{3422}{990}=3\frac{226}{495}\end{equation}$

You can also use your Casio classpad to do the conversion. Although I think it is easier just to do it yourself.

Let’s think about example 4,

$3.4\overline{56}=3.4+\frac{56}{1000}+\frac{56}{100000}+\frac{56}{10000000}+...$

Which is

$3.4+\frac{56}{1000\times 100^0}+\frac{56}{1000\times100^1}+\frac{56}{1000\times 100^2}...$

$3.4+\Sigma_{x=0}^\infty(\frac{56}{1000\times 100^x})$

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Filed under Arithmetic, Decimals, Fractions, Year 11 Specialist Mathematics

Tagged as recuring decimals to fractions, Year 11 Specialist Mathematics

May 15, 2025 · 10:01 am

Perfect Squares

Find all of the positive integers that make the following expression a perfect square.

(1) $\begin{equation*}(x-10)(x+14)\end{equation*}$

Let

$\begin{equation*}(x-10)(x+14)=n^2\end{equation}$

where $n$ is an integer.

Expand and simplify

$\begin{equation*}x^2+4x-140=n^2\end{equation}$

$\begin{equation*}x^2-4x-n^2=140\end{equation}$

Complete the square

$\begin{equation*}(x+2)^2-4-n^2=140\end{equation}$

$\begin{equation*}(x+2)^2-n^2=144\end{equation}$

Factorise (using difference of perfect squares)

$\begin{equation*}(x+2-n)(x+2+n)=144\end{equation}$

Find all of the factors of $144$

$(1,144), (2, 72), (3, 48), (4, 36), (6, 24), (8, 18), (9, 16), (12, 12)$

First pair,

$\begin{equation*}x+2-n=1 \tag {1} \end{equation}$

$\begin{equation*}x+2+n=144 \tag {2} \end{equation}$

$2x=141$

$x$ must be an integer.

I then used a spreadsheet

Hence the integers that make $(x-10)(x+14)$ are perfect square are, $10, 11, 13, 18,$ and $35$ .

Let’s try another one,

$(x-6)(x+14)$

(2) $\begin{equation*}(x-6)(x+14)=n^2\end{equation*}$

$\begin{equation*}(x^2+8x-84=n^2\end{equation}$

$\begin{equation*}(x+4)^2-n^2=100\end{equation}$

$\begin{equation*}(x+4-n)(x+4+n)=100\end{equation}$

Factors of 100,

$(1, 100), (2, 50), (4, 25), (5, 20), (10, 10)$

So the possible integers are $6$ and $22$ .

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Filed under Algebra, Arithmetic, Divisibility, Interesting Mathematics, Puzzles, Quadratic, Solving Equations

Tagged as perfect squares

September 23, 2024 · 3:11 am

Cats and Dogs

In my town 10% of the dogs think they are cats and 10% of the cats think they are dogs. All the other cats and dogs are perfectly normal. When all the cats and dogs in my town were rounded up and subjected to a rigorous test, 20% of them thought they were cats. What percentage of them really were cats?
Hamilton Olympiad 2003 B4 – The Ultimate Mathematical Challenge

Let $x$ be the number of cats and $y$ be the number of dogs.
Then $0.9x+0.1y$ think they are cats.
But we also know 20% of the total think they are cats.
$0.2(x+y)$
Therefore, $0.9x+0.1y=0.2(x+y)$
$0.9x+0.1y=0.2x+0.2y$
$0.7x=0.1y$
$7x=y$
Percentage of cats is $\frac{x}{x+y}\times100$
Substitute $7x$ for $y$
$\frac{x}{x+7x}\times100=\frac{x}{8x}\times100=\frac{1}{8}\times100=12.5%$
$\therefore 12.5$ % of the animals are cats

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Filed under Algebra, Arithmetic, Percentages, Simplifying fractions, UK Mathematics Challenge

Tagged as Fractions, percent, uk maths challenge

July 16, 2024 · 7:35 am

Australian Mathematics Competition – Polynomial Question

I came across this question from the 2010 Senior Australian Mathematics Competition:

A polynomial $f$ is given. All we know about it is that all its coefficients are non-negative integers, $f(1)=6$ and $f(7)=3438$ . What is the value of $f(3)$
Australian Mathematics Competition 2006-2012

I thought ‘excellent, a somewhat hard polynomial question for my students’ and then I tried it. Now I know why only 1% of students got it correct.

As we don’t know the order of the polynomial, let

$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0$

We know all of the coefficients are greater than or equal to zero. We also know

$f(1)=a_n+a_{n-1}+...+a+a_0=6$

Which means that all of the coefficients are between zero and six

$0\le{a_n}\le{6}$

We have also been given $f(7)$

$f(7)=7^na_n+7^{n-1}a_{n-1}+ ... +7a+1=3438$

As all of the coefficients are between zero and six, this is $3438$ written in base 7.

Let’s calculate a few powers of 7

	Powers of 7
$7^0$	1
$7^1$	7
$7^2$	49
$7^3$	343
$7^4$	2401
$7^5$	16807

Hence $3438$ written in base $7$ is $13011$

Therefore $f(x)=x^4+3x^3+x+1$

$f(3)=3^4+3\times3^3+3+1$

$f(3)=81+81+4$

$f(3)=166$

I really like this question. I think it could work well as a class extension activity with a bit of scaffolding.

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Filed under Number Bases, Polynomials

Tagged as base 7, number bases, Polynomials

May 12, 2024 · 11:50 am

Fractions to decimals

People usually know some fractions as decimals, for example

$\frac{1}{4}=0.25\ \textnormal{or }\frac{4}{5}=0.8$

And denominators that are powers of ten are also easy,

$\frac{47}{100}=0.47\ \textnormal{or }\frac{256}{1000}=0.256$

But what if it is something else? One that you don’t know. For example,

$\frac{5}{12}\ \textnormal{or }\frac{15}{37}$

I like to do these as a long division

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Filed under Arithmetic, Decimals, Fractions

Tagged as decimals, Fractions, fractions to decimals

April 14, 2024 · 6:35 am

Divisibility Rules – Integers 1 to 10

I think it’s useful to knot the divisibility rules, at least up to ten. Although, let’s face it, 7 is a bit tricky.

Divisibility Rules pdf

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Filed under Arithmetic, Divisibility

Tagged as Arithmetic, Divisibility rules

As numbers	As Powers of 7
$3438=1\times2401+1037$	$3438=1\times7^4+1037$
$1037=3\times343+8$	$1037=3\times7^3+8$
$8=1\times7+1$	$8=1\times7^1+1$
$1=1\times1$	$1=1\times7^0$

Category Archives: Arithmetic

Synthetic Division (for factorising and/or solving polynomials)

Converting recurring (non-terminating) decimals to fractions

Perfect Squares

Cats and Dogs

Australian Mathematics Competition – Polynomial Question

Fractions to decimals

Divisibility Rules – Integers 1 to 10

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