August 14, 2025 · 6:39 am

Trig Identities and Exact Values

My Year 11 Specialist Mathematics students are working on Trig identities. We came across this question

Without the use of a calculator, evaluate
(a) cos20^\circ\times cos40^\circ\times cos80^\circ

(b)cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})

OT Lee Year 11 Specialist Mathematics textbook

I spent a bit of time thinking about the question. Can you use a product to sum identity twice? But I was always being left with an angle that doesn’t have a nice exact value.

I tried a few things, had a chat to Meta AI, and finally stumbled upon this method.

Remember

    \begin{equation*}sin(2x)=2sin(x)cos(x)\end{equation}

Which can be rearranged to

    \begin{equation*}cos(x)=\frac{sin(2x)}{sin(x)}\end{equation}

(a) cos20^\circ\times cos40^\circ\times cos80^\circ=\frac{sin(40)}{2sin(20)}\frac{sin(80)}{2sin(40)}\frac{sin(160)}{2sin(80)}

Which simplifies to

    \begin{equation*}\frac{sin(160)}{8sin(20)}\end{equation}

Now sin(160)=sin(20)

Hence cos20^\circ\times cos40^\circ\times cos80^\circ=\frac{1}{8}

And we will do the same for part (b)

cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})=\frac{sin(\frac{2\pi}{7})}{2sin(\frac{\pi}{7})}\frac{sin(\frac{4\pi}{7})}{2sin(\frac{2\pi}{7})}\frac{sin(\frac{8\pi}{7})}{2sin(\frac{4\pi}{7})}

Which simplifies to

cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})=\frac{sin(\frac{8\pi}{7})}{8sin(\frac{\pi}{7})}

Now sin(\frac{8\pi}{7})=-sin(\frac{\pi}{7})

Hence cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})=\frac{-1}{8}

And then I had to test them on my Classpad.

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Filed under Classpad Skills, Identities, Simplifying fractions, Trigonometry, Year 11 Specialist Mathematics

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August 10, 2025 · 4:54 am

Trigonometric Identities – Product to Sum

Let’s think about the sine and cosine addition and subtraction trig identities.

(1)   \begin{equation*}sin(A+B)=sinAcosB+cosAsinB\end{equation*}

(2)   \begin{equation*}sin(A-B)=sinAcosB-cosAsinB\end{equation*}

If we add equation 1 and 2, we get

    \begin{equation*}sin(A+B)+sin(A-B)=2sinAcosB\end{equation}

Hence, sinAcosB=\frac{1}{2}(sin(A+B)+sin(A-B))

If we subtract equation 2 from equation 1, we get

    \begin{equation*}sin(A+B)-sin(A-B)=2cosAsinB\end{equation}

Hence, cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B)

What about the cosine addition and subtraction idenities?

(3)   \begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation*}

(4)   \begin{equation*}cos(A-B)=cosAcosB+sinAsinB\end{equation*}

If we add equation 3 and 4, we get

    \begin{equation*}cos(A+B)+cos(A-B)=2cosAcosB\end{equation}

Hence, cosAcosB=\frac{1}{2}(cos(A+B)+cos(A-B))

If we subtract 3 from 4, we get

    \begin{equation*}cos(A-B)-cos(A+B)=2sinAsinB\end{equation}

Hence, sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))

These are the product to sum identities.

    \begin{equation*}sinAcosB=\frac{1}{2}(sin(A+B)+sin(A-B))\end{equation}


    \begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}


    \begin{equation*}cosAcosB=\frac{1}{2}(cos(A+B)+cos(A-B))\end{equation}


    \begin{equation*}sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))\end{equation}

Examples

(1) Solve sin(5x)-sin(x)=0 for 0\le x \le 2\pi

Remember,

    \begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}

    \begin{equation*}A+B=5\end{equation}

    \begin{equation*}A-B=1\end{equation}

Therefore, A=3 and B=2

    \begin{equation*}sin(5x)-sin(x)=2cos(3x)sin(2x)=0\end{equation}

    \begin{equation*}cos(3x)=0\end{equation}

    \begin{equation*}3x=\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{\9\pi}{2}, \frac{11\pi}{2}\end{equation}

    \begin{equation*}x=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\end{equation}

    \begin{equation*}sin(2x)=0\end{equation}

    \begin{equation*}2x=0, \pi, 2\pi, 3\pi, 4\pi\end{equation}

    \begin{equation*}x=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\end{equation}

Hence x=0, \frac{\pi}{6}. \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}, 2\pi

(2)Solve sin(7\theta)-sin(\theta)=sin(3\theta) for 0 \le \theta \le\2\pi

    \begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}

Therefore, A+B=7 and A-B=1

A=4, B=3

    \begin{equation*}2cos(4\theta)sin(3\theta)=sin(3\theta)\end{equation}

    \begin{equation*}2cos(4\theta)sin(3\theta)-sin(3\theta)=0\end{equation}

    \begin{equation*}sin(3\theta)(2cos(4\theta)-1)=0\end{equation}

sin(3\theta)=0 and cos(4\theta)=\frac{1}{2}

3\theta=0, \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi

\theta=0, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}. 2\pi

cos(4\theta)=\frac{1}{2}

4\theta=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3}, \frac{19\pi}{3}, \frac{23\pi}{3}

\theta=\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}, \frac{23\pi}{12}

Hence \theta=0, \frac{\pi}{3}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{5\pi}{3}, \frac{23\pi}{12}, 2\pi

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August 2, 2025 · 9:30 am

Trigonometric Exact Value

Using an appropriate double angle identity, find the exact value of
cos(\frac{\pi}{12})

The double angle identity for sine is

(1)   \begin{equation*}cos(2A)=cos^2A-sin^2A=2cos^2A-1=1-2sin^2A\end{equation*}

That means \frac{\pi}{12} is either 2A or A.

It must be A as 2\times\frac{\pi}{12}=\frac{\pi}{6} as there are exact values for \frac{\pi}{6}

Hence,

    \begin{equation*}cos{\frac{\pi}{6}}=2cos^2{\frac{\pi}{12}}-1\end{equation}

    \begin{equation*}\frac{\sqrt{3}}{2}=2cos^2{\frac{\pi}{12}}-1\end{equation}

    \begin{equation*}\frac{\sqrt{3}}{2}+1=2cos^2{\frac{\pi}{12}}\end{equation}

    \begin{equation*}\frac{\frac{\sqrt{3}}{2}+1}{2}=cos^2{\frac{\pi}{12}}\end{equation}

    \begin{equation*}\frac{\sqrt{3}+2}{4}=cos^2{\frac{\pi}{12}}\end{equation}

    \begin{equation*}\sqrt{\frac{\sqrt{3}+2}{4}}=cos{\frac{\pi}{12}}\end{equation}

As \frac{\pi}{12} is in the first quadrant, we don’t need to consider the negative version.

    \begin{equation*}cos(\frac{\pi}{12})=\frac{\sqrt{3}+2}{2}\end{equation}

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July 28, 2025 · 8:09 am

Trig Identities – Addition and Subtraction

Deriving the addition and subtraction trigonometric identities.

We will start with cosine, and use the result to derive the remaining identities.

Proving cos(A-B)=cos(A)cos(B)+sin(A)(sin(B).

A and B are represented in the unit circle below.

Remember P(x_1,y_1)=(cos(A), sin(A)) and Q(x_2, y_2)=(cos(B), sin(B))

Using the cosine rule and triangle OPQ, find PQ

    \begin{equation*}(PQ)^2=1^2+1^2-2(1)(1)cos(A-B)\end{equation}

    \begin{equation*}(PQ)^2=2-2cos(A-B)\end{equation}

Using the distance between points, find PQ

    \begin{equation*}(PQ)^2=(x_1-x_2)^2+(y_1-y_2)^2\end{equation}

    \begin{equation*}(PQ)^2=(cosA-cosB)^2+(sinA-sinB)^2\end{equation}

(PQ)^2=cos^2A-2cosAcosB+cos^2B+sin^2A-2sinAsinB+sin^2B

Remember the Pythagorean identity

    \begin{equation*}cos^2\theta+sin^2\theta=1\end{equation}

    \begin{equation*}(PQ)^2=2-2cosAcosB-2sinAsinB\end{equation}

Hence

    \begin{equation*}2-2cos(A-B)=2-2cosAcosB-2sinAsinB\end{equation}

(1)   \begin{equation*}cos(A-B)=cosAcosB+sinAsinB\end{equation*}

We can then use this identity to find cos(A+B).

    \begin{equation*}cos(A+B)=cos(A-(-B))\end{equation}

    \begin{equation*}cos(A-(-B))=cos(A)cos(-B)+sinAsin(-B)\end{equation}

Remember cos(-B)=cos(B) and sin(-B)=-sin(B)

    \begin{equation*}cos(A-(-B))=cosAcosB-sinAsinB\end{equation}

(2)   \begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation*}

We can also find sin(A+B)

Remember, sin\theta=cos(\frac{\pi}{2}-\theta)

    \begin{equation*}sin(A+B)=cos(\frac{\pi}{2}-(A+B))\end{equation}

    \begin{equation*}sin(A+B)=cos((\frac{\pi}{2}-A)-B)\end{equation}

    \begin{equation*}sin(A+B)=cos(\frac{\pi}{2}-A)cosB+sin(\frac{\pi}{2}-A)sinB\end{equation}

(3)   \begin{equation*}sin(A+B)=sinAcosB+cosAsinB\end{equation*}

We can use equation 2 to find sin(A-B)

    \begin{equation*}sin(A-B)=sin(A+(-B))\end{equation}

    \begin{equation*}sin(A+(-B))=sinAcos(-B)+cosAsin(-B)\end{equation}

    \begin{equation*}sin(A+(-B))=sinAcos(B)-cosAsin(B)\end{equation}

(4)   \begin{equation*}sin(A-B)=sinAcos(B)-cosAsin(B)\end{equation*}

And we can use both the sine and cosine identities to find tan(A+B)

Remember tan\theta=\frac{sin\theta}{cos\theta}

    \begin{equation*}tan(A+B)=\frac{sin(A+B)}{cos(A+B)}\end{equation}

    \begin{equation*}=\frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}\end{equation}

    \begin{equation*}=\frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}\times \frac{cosAcosB}{cosAcosB}\end{equation}

    \begin{equation*}=\frac{\frac{sinAcosB}{cosAcosB}+\frac{cosAsinB}{cosAcosB}}{\frac{cosAcosB}{cosAcosB}+\frac{sinAsinB}{cosAcosB}}\end{equation}

    \begin{equation*}=\frac{tanA+tanB}{1-tanAtanB}\end{equation}

(5)   \begin{equation*}tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}\end{equation*}

and

(6)   \begin{equation*}tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}\end{equation*}

    \begin{equation*}cos(A\pm B)=cosAcosB\mp sinAsinB\end{equation}


    \begin{equation*}sin(A\pm B)=sinAcosB\pm cosAsinB\end{equation}


    \begin{equation*}tan(A \pm B)=\frac{tan A \pm tanB}{1 \mp tanAtanB}\end{equation}

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Filed under Addition and Subtraction Identities, Identities, Non-Right Trigonometry, Simplifying fractions, Trigonometry, Year 11 Mathematical Methods, Year 11 Specialist Mathematics

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July 26, 2025 · 8:45 am

General Solutions to Trigonometric Equations

Solve sinx=\frac{1}{2} for 0\le x \le 2\pi

Sine is positive in the first and second quadrants.

    \begin{equation*}sinx=\frac{1}{2}\end{equation}

    \begin{equation*}x=\frac{\pi}{6} \text{ and } x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\end{equation}

But what if we aren’t given a domain for the x values?

Then we need to give general solutions.

For example,

Solve sinx=\frac{1}{2}

As you can see from the sketch above, there are infinite solutions.

The sine function has a period of 360^\circ, and so if \frac{\pi}{6} is a solution then \2pi+\frac{\pi}{6} is also a solution. This means \frac{\pi}{6}+2\pi n, n\in\mathbb{Z} is a general solution. And we can do the same for the second solution \frac{5\pi}{6}+2\pi n.

In general

    \begin{equation*}sinx=y\end{equation}


    \begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi-arcsin(y)+2\pi n \end{equation}


    \begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi(2n+1)-arcsin(y), n \in \mathbb{Z}\end{equation}


We can turn this into one equation

    \begin{equation*}x=(-1)^n arcsin(y)+n\pi, n \in \mathbb{Z}\end{equation}

What about cosine?

Solve cosx=\frac{1}{2}

Cosine is positive in the first and fourth quadrants (it also has a period of 2\pi. The first two (positive) solutions are \frac{\pi}{3} and 2\pi-\frac{\pi}{3}.

To generalise, x=2\pi n+\frac{\pi}{3} \text { and }x=2\pi n -\frac{\pi}{3}, which we can make into one equation x=2\pi n \pm \frac{pi}{3}

In general

    \begin{equation*}cosx=y\end{equation}

    \begin{equation*}x=2\pi n \pm arccos(y), n\in\mathbb{Z}\end{equation}

What about the tangent function? Remember tan has a period of \pi.

Solve tanx=\sqrt{3}

First, note that the solutions are all a common distance (\pi) apart.

Tan is positive in the first and the third quadrant

    \begin{equation*}tanx=\sqrt{3}\end{equation}

    \begin{equation*}x=\frac{\pi}{3} \text { and } x=\pi+\frac{\pi}{3}\end{equation}

Because all of the solutions are \pi radians apart, the general solution is x=\frac{\pi}{3} \pm \pi

In general

    \begin{equation*}tanx=y\end{equation}

    \begin{equation*}x=arctan(y) + n\pi, n\in \mathbb{Z}\end{equation}

Examples

Solve for all values of x, tan^2(x)+tan(x)-6=0

    \begin{equation*}tan^2(x)+tan(x)-6=0\end{equation}

This is a quadratic equation – we need two numbers that add to 1 and multiple to -6, +3 \text { and } -2

    \begin{equation*}(tan(x)+3)(tan(x)-2))=0\end{equation}

    \begin{equation*}tan(x)=-3 \text { or } tan(x)=2\end{equation}

    \begin{equation*}x=arctan(-3)+n\pi \text { or } x=arctan(2)+n\pi, n\in\mathbb{Z}\end{equation}

Solve 2cos(2x+\frac{\pi}{18})=\sqrt{3}

    \begin{equation*}2cos(2x+\frac{\pi}{18})=\sqrt{3}\end{equation}

    \begin{equation*}cos(2x+\frac{\pi}{18})=\frac{\sqrt{3}}{2}\end{equation}

    \begin{equation*}2x+\frac{\pi}{18}=2n\pi \pm \frac{\pi}{6}\end{equation}

    \begin{equation*}2x=2n\pi \pm \frac{\pi}{6}-\frac{\pi}{18}\end{equation}

    \begin{equation*}2x=2n\pi \pm \frac{\pi}{9}\end{equation}

    \begin{equation*}x=n\pi \pm \frac{\pi}{18}\end{equation}

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Filed under Algebra, Quadratic, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

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July 22, 2025 · 6:43 am

Continuous Uniform Random Variable

My Year 12 Mathematics Methods students are doing continuous random variables at the moment and I thought it would be worthwhile deriving the mean and variance formulas for a uniform continuous random variable.

The probability density function for a uniform random variable is

    \begin{equation*}f(x)= \left \{ {\begin{matrix}\frac{1}{b-a} & a\le x\le b \\0 & \text {elsewhere}\end{matrix}}\end{equation}

and it looks like

Remember, the mean \mu or expected value E(X) of a continuous random variable is

(1)   \begin{equation*}E(X)=\int xp(x) dx\end{equation*}

and the variance \sigma^2 is

(2)   \begin{equation*}\sigma^2=\int (x-\mu)^2p(x) dx\end{equation*}

We are going to use equations 1 and 2 to find formulae for a uniform continuous random variable.

    \begin{equation*}\mu=\int_a^b x (\frac{1}{b-a}) dx\end{equation}

    \begin{equation*}\mu=\frac{x^2}{2(b-a)}|\begin{matrix}b\\a\end{matrix}\end{equation}

    \begin{equation*}\mu=\frac{b^2}{2(b-a)}-\frac{a^2}{2(b-a)}=\frac{b^2-a^2}{2(b-a)}\end{equation}

Factorise the numerator (using difference of squares)

    \begin{equation*}\mu=\frac{(b-a)(b+a)}{2(b-a)}\end{equation}

Hence,

    \begin{equation*}\mu=\frac{b+a}{2}\end{equation}

Now for the variance

    \begin{equation*}\sigma^2=\int_a^b (x-(\frac{a+b}{2}))^2(\frac{1}{b-a}) dx\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{(x-\frac{a+b}{2})^3}{3})|\begin{matrix}b\\a\end{matrix}\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}((\frac{(b-\frac{a+b}{2})^3}{3})-(\frac{(a-\frac{a+b}{2})^3}{3}))\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{-a^3}{12}+\frac{b^3}{12}+\frac{a^2b}{4}-\frac{ab^2}{4})\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{-a^3+b^3+3a^2b-3ab^2}{12})\end{equation}

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{b^3-3b^2a+3ba^2-a^3}{12})\end{equation}

From the binomial expansion theorem, we know

    \begin{equation*}b^3-3b^2a+3ba^2-a^3=(b-a)^3\end{equation}

Hence

    \begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{(b-a)^3}{12}\end{equation}

and

    \begin{equation*}\sigma^2=\frac{(b-a)^2}{12}\end{equation}

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July 20, 2025 · 5:50 am

Integrating the Natural Log Function

One of the students asked me the other day how to integrate f(x)=ln(x) – it’s not part of their course, but I thought I would do it here.

We use integration by parts to integrate ln(x)

    \begin{equation*}\int u dv=uv-\int v du\end{equation}

    \begin{equation*}\int ln(x) dx\end{equation}

Let u=ln(x) and dv=1, then du=\frac{1}{x} and v=x

    \begin{equation*}\int ln(x) dx = xln(x)-\int x\times \frac{dx}{x}\end{equation}

    \begin{equation*}\int ln(x) dx = xln(x)-\int 1 dx\end{equation}

    \begin{equation*}\int ln(x) dx = xln(x)-x+c\end{equation}

What about \int 5ln(\sqrt{x}) dx

We can take advantage of log laws and the properties of integration.

    \begin{equation*}\int 5ln(\sqrt{x}) dx=5\int lnx^{\frac{1}{2}} dx=5\int \frac{1}{2}ln(x) dx=\frac{5}{2} \int ln(x) dx\end{equation}

    \begin{equation*}\frac{5}{2} \int ln(x) dx=\frac{5}{2}(xlnx-x)+c\end{equation}

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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

\alpha+\beta+\gamma=\frac{-b}{a}

The product of the roots

\alpha\beta\gamma=\frac{-d}{a}

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

21-1031-30
2-1630
1-8150

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.

Factors of ThirtyX+Y+ZXY+YZ+XZ
1, 1, 301+1+30=321+30+30=61
1, 2, 151+2+15=182+15+30=47
1, 5, 61+5+6=125+6+30=41
2, 3, 52+3+5=106+10+15=31

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

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

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

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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.

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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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