Category Archives: Algebra

July 11, 2025 · 5:35 am

Simultaneous Equation (or is it?)

Solve simultaneously

$X+Y+Z=10$
$XYZ=30$
$XY+YZ+XZ=31$

We could attempt to solve this simultaneously, but I think the algebra would be tricky.

The three equations are related to the roots of a cubic polynomial.

If the general equation of the polynomial is $ax^3+bx^2+cx+d$ , then we know

The sum of the roots

The product of the roots

and

\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}

So from our three equations we have

(1) $\begin{equation*}X+Y+Z=10=\frac{-b}{a}\end{equation*}$

(2) $\begin{equation*}XYZ=30=\frac{-d}{a}\end{equation*}$

(3) $\begin{equation*}XY+YZ+XZ=31=\frac{c}{a}\end{equation*}$

Let $a=1$ , then $b=-10, c=31$ , and $d=-30$

Our cubic is $t^3-10t^2+31t-30$ and we can try to solve it.

The roots will be factors of $-30$ , so $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$

Try $t=2$

$\begin{equation*}(2)^3-10(2)^2+31(2)-30=0\end{equation}$

Hence $t=2$ is a root.

Use synthetic division to find the quadratic factor

$2$	$1$	$-10$	$31$	$-30$
		$2$	$-16$	$30$
	$1$	$-8$	$15$	$0$

The quadratic factor is $x^2-8x+15$ , which factorises to $(x-3)(x-5)$

Hence the solutions are $X=2, Y=3$ , and $Z=5$

We could assume the solutions are natural numbers, then we can look at factors of 30.

Hence the solutions are $X=2, Y=3,$ and $Z=5$

But with this approach we might not be able to find the solutions.

Leave a Comment

Filed under Algebra, Cubics, Factorising, Polynomials, Quadratic, Simultaneous Equations, Solving, Sum and Product of Roots

Tagged as simultaneous equations, solving cubic equations, sum and product of roots

July 8, 2025 · 6:17 am

Age Question (Year 8 equation solving)

Eight years ago my father was three times as old as I shall be in five years time. When I was born he was 41 years old. How old am I now?

I always find these age questions a bit weird – a bit of a riddle, and contrived (just so we can solve some equations)

Let $x$ be my age now, and $y$ be my fathers age now.

(1) $\begin{equation*}y=x+41\end{equation*}$

Because my father was 41 when I was born.

(2) $\begin{equation*}y-8=3(x+5)\end{equation*}$

$y-8$ for 8 years ago, and $3(x+5)$ for three times my age in 5 years.

Solve the equations simultaneously. Substitute $y=x+41$ into equation $2$

$\begin{equation*}x+41-8=3(x+5)\end{equation}$

$\begin{equation*}x-33=3x+15\end{equation}$

$\begin{equation*}18=2x\end{equation}$

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

Hence my current age is $9$ .

Leave a Comment

Filed under Algebra, Simultaneous Equations, Solving Equations, Year 8 Mathematics

Tagged as age question, simultaneous equations, Solving Equations

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.

Leave a Comment

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

Tagged as non monic synthetic division, Synthetic division

June 26, 2025 · 7:22 am

Trigonometry Question

I don’t know where I found this question, but it does require algebra and problem solving (as well as right trig and Pythagoras)

From a point $A$ , a lighthouse is on a bearing of $026^\circ$ T and the top of the light house is at an angle of elevation of $20.25^\circ$ . From a point $B$ , the lighthouse is on a bearing of $296^\cric$ T and the top of the lighthouse is at angle of elevation of $10.2^\circ$ . If $A$ and $B$ are 500 metres apart, find the height of the lighthouse.

Let’s draw a diagram.

Let the height of the lighthouse be $h$

We can find the angle between $A$ , the lighthouse, and $B$ by using the base triangle

The red line from $L$ is parallel to the two north lines. Hence $\theta=26^\circ+64^\circ=90^\circ$ (Alternate angles in parallel lines are congruent)

It’s a right triangle so we know

(1) $\begin{equation*}500^2=(AL)^2+(BL)^2\end{equation*}$

We are going to use the other two triangles to find $AL$ and $BL$

Substitute $AL$ and $BL$ into equation $1$

$\begin{equation*}500^2=(\frac{h}{tan(20.25)})^2+(\frac{h}{tan(10.2)})^2\end{equation}$

Solve for $h$ .

$\begin{equation*}500^2=7.35h^2+30.89h^2=38.24h^2\end{equation}$

$\begin{equation*}h^2=6538.3\end{equation}$

$\begin{equation*}h=80.9m\end{equation}$

Leave a Comment

Filed under Algebra, Bearings, Pythagoras, Right Trigonometry, Solving Equations, Trigonometry

Tagged as hard trig, trigonometry question

June 10, 2025 · 1:54 am

Square Root Puzzle

Is it possible to find three numbers, $a, b, c$ , none of which is zero or a perfect square, such that
$\sqrt{a}+\sqrt{b}=\sqrt{c}$

Can You Solve These – David Wells

As $a, b$ and $c$ can’t be perfect squares, let $a=d\times e^2, b=f\times g^2$ and $c=h\times k^2$ where $d, e, f, g, h$ and $k$ are real numbers.

Hence $\sqrt{a}=e\sqrt{d}, \sqrt{b}=g\sqrt{f}$ and $\sqrt{c}=k\sqrt{h}$ .

$\begin{equation*}\sqrt{a}+\sqrt{b}=\sqrt{c}\end{equation}$

$\begin{equation*}e\sqrt{d}+g\sqrt{f}=k\sqrt{h}\end{equation}$

For the above equation to be possible $d, f$ and $h$ must simplify to the same surd. Because we are looking for one set of numbers, let $d=f=h$ .

$\begin{equation*}e\sqrt{d}+g\sqrt{d}=k\sqrt{d}\end{equation}$

$\begin{equation*}e+g=k\end{equation}$

Let’s think of some numbers that might work…

$1+2=3$ or $2+3=5$ , etc.

Let’s try $e=1, g=2,$ and $k=3$

We now have $a=d, b=4d,$ and $c=9d$

As $a$ can’t be a square number, $d$ can’t be a square number.

Try $d=2$

$\begin{equation*}\sqrt{2}+\sqrt{8}=\sqrt{18}\end{equation}$

$LHS=\sqrt{2}+2\sqrt{2}$

$LHS=3\sqrt{2}$

$LHS=\sqrt{9\times 2}$

$LHS=\sqrt{18}$

$LHS=RHS$

One set of possible numbers are $2, 8,$ and $18$ .

Leave a Comment

Filed under Algebra, Interesting Mathematics, Puzzles

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 $22$ .

Leave a Comment

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

Tagged as perfect squares

May 6, 2025 · 7:27 am

Combinations Proof

Prove $\begin{pmatrix}n+1\\r\end{pmatrix}=\begin{pmatrix}n\\r-1\end{pmatrix}+\begin{pmatrix}n\\r\end{pmatrix}$

Let’s start with the right hand side.

$RHS=\begin{pmatrix}n\\r-1\end{pmatrix}+\begin{pmatrix}n\\r\end{pmatrix}$

$RHS=\frac{n!}{(n-(r-1))!(r-1)!}+\frac{n!}{(n-r)!r!}$

Simplify

$RHS=\frac{n!}{(n+1-r)!}+\frac{n!}{(n-r)!r!}$

There is a common denominator of $(n+1-r)!r!$

$RHS=\frac{n!}{(n+1-r)!}\times \frac{r}{r}+\frac{n!}{(n-r)!r!}\times \frac{n+1-r}{n+1-r}$

$RHS=\frac{n!r+n!(n+1-r)}{(n+1-r)r!}$

$RHS=\frac{n!r+(n+1)n!-n!r}{(n+1-r)!r!}$

$RHS=\frac{(n+1)!}{(n+1-r)!r!}$

$RHS=\begin{pmatrix}n+1\\r\end{pmatrix}$

$RHS=LHS$

We know this intuitively from Pascal’s triangle

Where each entry is the sum of the two entries above it – for example,

$6+4=10$

Remember, Pascal’s triangle can be written as combinations,

so $\begin{pmatrix}5\\3\end{pmatrix}=\begin{pmatrix}4\\2\end{pmatrix}+\begin{pmatrix}4\\3\end{pmatrix}$

Leave a Comment

Filed under Algebra, Counting Techniques, Year 11 Mathematical Methods

May 3, 2025 · 3:34 am

Geometry Puzzle

Another puzzle from this book

Two ladders are propped up vertically in a narrow passageway between two vertical buildings. The ends of the ladders are 8 metres and 4 metres above the pavement.
Find the height above the ground, $T$ ,

$\Delta ABC \sim \Delta TEC$ and $\Delta DCB \sim \Delta TEB$ (Angle Angle Similarity)

Hence,

(1) $\begin{equation*}\frac{h}{8}=\frac{x}{x+y}\end{equation*}$

(2) $\begin{equation*}\frac{h}{4}=\frac{y}{x+y}\end{equation*}$

From equation $1$ $h=\frac{8x}{x+y}$ and from $2$ $h=\frac{4y}{x+y}$

Hence, $\frac{8x}{x+y}=\frac{4y}{x+y}$

Therefore, $8x=4y$ and $y=2x$

From equation $1$

$\begin{equation*}h=\frac{8x}{3x}=\frac{8}{3}\end{equation}$

Hence $T$ is $2 \frac{2}{3}$ m above the ground.

Leave a Comment

Filed under Geometry, Puzzles, Similarity, Simplifying fractions, Solving Equations

Tagged as Puzzle

April 30, 2025 · 6:08 am

Factorising Non-Monic Quadratics

The general equation of a quadratic is $ax^2+bx+c$

Let’s explore different methods of factorising a non-monic quadratic (the $a$ term is not $1$ )

Factorise $6x^2+x-12$

We need to find two numbers that add to $1$ and multiply to $-72$ (i.e. add to $b$ and multiply to $a\times c)$

The two numbers are $9$ and $-8$

Method 1 – Splitting the middle term

This is the method I teach the most often

$\begin{equation*}6x^2+x-12\end{equation}$

Split the middle term (the $b$ term) into the two numbers

$\begin{equation*}6x^2-8x+9x-12\end{equation}$

The order doesn’t matter.

Find a common factor for the first term terms, and then for the last two terms.

$\begin{equation*}2x(3x-4)+3(3x-4)\end{equation}$

There is a common factor of $(3x-4)$ , factor it out.

$\begin{equation*}(3x-4)(2x+3)\end{equation}$

Method two – Fraction

$\begin{equation*}6x^2+x-12\end{equation}$

Put $6x$ into both factors and divide by $6$

$\begin{equation*}\frac{(6x-8)(6x+9)}{6}\end{equation}$

Factorise

$\begin{equation*}\frac{2(3x-4)3(2x+3)}{6}\end{equation}$

$\begin{equation*}\frac{6(3x-4)(2x+3)}{6}\end{equation}$

$\begin{equation*}(3x-4)(2x+3)\end{equation}$

Method 3 – Monic to non-monic

$\begin{equation*}y=6x^2+x-12\end{equation}$

Multiply both sides of the equation by $a$

$\begin{equation*}6y=6(6x^2+x-12)\end{equation}$

$\begin{equation*}6y=6^2x^2+6x-72\end{equation}$

$\begin{equation*}6y=(6x)^2+6x-72\end{equation}$

Let $A=6x$

$\begin{equation*}6y=A^2+A-72\end{equation}$

Factorise

$\begin{equation*}6y=(A+9)(A-8)\end{equation}$

Replace the $A$ with $6x$

$\begin{equation*}6y=(6x+9)(6x-8)\end{equation}$

$\begin{equation*}6y=3(2x+3)2(3x-4)\end{equation}$

$\begin{equation*}6y=6(2x+3)(3x-4)\end{equation}$

$\begin{equation*}y=(2x+3)(3x-4)\end{equation}$

Method 4 – Cross Method

$\begin{equation*}6x^2+x-12\end{equation}$

Place the two numbers in the cross

Place the two numbers that add to $1$ and multiply to $-72$ in the other parts of the cross.

Divide these two numbers by $6$ (i.e $a$ )

Simplify

Hence,

$\begin{equation*}(x-\frac{4}{3})(x+\frac{3}{2})\end{equation}$

Which is

$\begin{equation*}(3x-4)(2x+3)\end{equation}$

Method 5 – By Inspection

This is my least favourite method – although students get better with practice

$\begin{equation*}6x^2+x-12\end{equation}$

The factors of $a$ are $1, 2, 3,$ and $6$ and the factors of $12$ are $1, 2, 3, 4, 6, 12$

We know one number is positive and one number negative.

Which give us all of these possibilities

$\begin{equation*}(2x+3)(3x-4)\end{equation}$

With a bit of practice you don’t need to check all of the possibilties, but I find students struggle with this method.

Method 6 – Grid

$\begin{equation*}6x^2+x-12\end{equation}$

Create a grid like the one below


	$6x^2$
		$-12$

Find the two numbers that multiply to $-72$ and add to $1$ and place them in the other grid spots (see below)


	$6x^2$	$-8x$
	$9x$	$-12$

Find the HCF (highest common factor) of each row and put in the first column.

Row $1$ HCF= $2x$ , Row $2$ HCF= $3$


$2x$	$6x^2$	$-8x$
$3$	$9x$	$-12$

For the columns, calculate what is required to multiple the HCF to get the table entry.

For example, what do you need to multiple $2x$ and $3$ by to get $6x^2$ and $9x$ ? In this case it is $3x$ . It’s always going to be the same thing, so just use one value to calculate it,

	$3x$	$-4$
$2x$	$6x^2$	$-8x$
$3$	$9x$	$-12$

The factors are column $1$ and row $1$
$(2x+3)(3x-4)$

The two methods I use the most are splitting the middle term, and the cross method, but I can see value in the grid method.

Leave a Comment

Filed under Algebra, Factorising, Factorising, Polynomials, Quadratic, Quadratics

Tagged as Factorising, factorising non-monic trinomials, non-monic quadratics

April 16, 2025 · 8:26 am

Binomial Expansion – deriving the formula for Variance

We have seen how the formula for mean (expected value) was derived, and now we are going to look at variance.

In general variance of a probability distribution is

(1) $\begin{equation*} Var(X)=E(X^2)-(E(X))^2\end{equation*}$

We are going to start by calculating $E(X^2)$

$\begin{equation*}E(X^2)=\sum_{x=0}^nx^2p(x)\end{equation}$

$\begin{equation*}E(X^2)=\sum_{x=0}^nx^2\begin{pmatrix}n\\x\end{pmatrix}p^x(1-p)^{n-x}\end{equation}$

$\begin{equation*}E(X^2)=\sum_{x=0}^n x^2\frac{n!}{(n-x)!x!}p^x(1-p)^{n-x}\end{equation}$

The $x^2$ cancels with the $x!$ to leave $x$ on the numerator and $(x-1)!$ on the denominator.

Also, when $x=0, x^2=0$ and we can start the sum at $x=1$

$\begin{equation*}E(X^2)=\sum_{x=1}^n x\frac{n!}{(n-x)!(x-1)}!p^x(1-p)^{n-x}\end{equation}$

Let $y=x-1$ and $m=n-1$ , when $x=n, y=n-1$ and hence $y=m$ and when $x=1, y=0$

Our equation is now

$\begin{equation*}E(X^2)=\sum_{y=0}^m (y+1)\frac{(m+1)!}{(m+1-(y+1))!y!}p^{y+1}(1-p)^{m+1-(y+1)}\end{equation}$

Simplify

$\begin{equation*}E(X^2)=\sum_{y=0}^m(y+1)\frac{(m+1)!}{(m-y)!y!}p^{y+1}(1-p)^{m-y}\end{equation}$

$\begin{equation*}E(X^2)=\sum_{y=0}^m(y+1)(m+1)\frac{m!}{(m-y)!y!}p\times p^y(1-p)^{m-y}\end{equation}$

$\begin{equation*}E(X^2)=\sum_{y=0}^m p(y+1)(m+1)\frac{m!}{(m-y)!y!} p^y(1-p)^{m-y}\end{equation}$

$\begin{equation*}E(X^2)=\sum_{y=0}^m p(y+1)(m+1)p(y)\end{equation}$

$\begin{equation*}E(X^2)=p(m+1)(\sum_{y=0}^myp(y)+\sum_{y=0}^mp(y))\end{equation}$

$\sum_{y=0}^mp(y)=1$ and $\sum_{y=0}^myp(y)=E(Y)$

$\begin{equation*}E(X^2)=p(m+1)(E(Y)+1)\end{equation}$

$E(Y)=mp$

$\begin{equation*}E(X^2)=p(m+1)(mp+1)\end{equation}$

$m+1=n$

$\begin{equation*}E(X^2)=pn((n-1)p+1)\end{equation}$

$\begin{equation*}E(X^2)=n^2p^2-np^2+np\end{equation}$

Now from equation $1$

$\begin{equation*}Var(X)=E(X^2))-(E(X))^2\end{equation}$

$\begin{equation*}=Var(X)=n^2p^2-np^2+np-n^2p^2\end{equation}$

$\begin{equation*}Var(X)=-np^2+np\end{equation}$

(2) $\begin{equation*}Var(X)=np(1-p)\end{equation*}$

and the standard deviation is

(3) $\begin{equation*}\sigma_X=\sqrt{np(1-p)}\end{equation*}$

1 Comment

Filed under Algebra, Binomial, Probability Distributions, Standard Deviation

Tagged as binomial distribution, standard deviation of a binomial distribution, variance of a binomial distribution

Factors of Thirty	$X+Y+Z$	$XY+YZ+XZ$
$1, 1, 30$	$1+1+30=32$	$1+30+30=61$
$1, 2, 15$	$1+2+15=18$	$2+15+30=47$
$1, 5, 6$	$1+5+6=12$	$5+6+30=41$
$2, 3, 5$	$2+3+5=10$	$6+10+15=31$

Possible factorisations	$b$ term of expansion
$(x-1)(6x+12)$	$12x-6x=6x$	No
$(x-2)(6x+6)$	$6x-12x=-6x$	No
$(x-3)(6x+4)$	$4x-18x-14x$	No
$(2x-1)(3x+12)$	$24x-3x=21x$	No
$(2x-2)(3x+6)$	$12x-6x=6x$	No
$(2x-3)(3x+4)$	$8x-9x=-1x$	Almost, switch the signs
$(2x+3)((3x-4)$	$-8x+9x=1x$	Yes

Category Archives: Algebra

Simultaneous Equation (or is it?)

Age Question (Year 8 equation solving)

Synthetic Division (for factorising and/or solving polynomials)

Trigonometry Question

Square Root Puzzle

Perfect Squares

Combinations Proof

Geometry Puzzle

Factorising Non-Monic Quadratics

Method 1 – Splitting the middle term

Method two – Fraction

Method 3 – Monic to non-monic

Method 4 – Cross Method

Method 5 – By Inspection

Method 6 – Grid

Binomial Expansion – deriving the formula for Variance

Recent Posts

Categories

Archives