Category Archives: Polynomials

March 31, 2025 · 7:14 am

Deriving the Quadratic Equation formula

My year 10 students have been learning how to complete the square with the idea of then deriving the quadratic equation formula.

The general equation for a quadratic is y=ax^2+bx+c

Completing the square,

    \begin{equation*}ax^2+bx+c\end{equation}

Factorise out the leading coefficient (i.e. a)

    \begin{equation*}a(x^2+\frac{bx}{a}+\frac{c}{a})\end{equation}

Half the second term (i.e \frac{b}{a}) and subtract the square of the second term.

    \begin{equation*}a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a})\end{equation}

    \begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a})\end{equation}

Simplify

    \begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})\end{equation}

    \begin{equation*}a((x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a^2})\end{equation}

    \begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}\end{equation}

Now let’s solve

    \begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}=0\end{equation}

    \begin{equation*}a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}\end{equation}

    \begin{equation*}(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\end{equation}

    \begin{equation*}(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\end{equation}

    \begin{equation*}(x+\frac{b}{2a})=\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}

    \begin{equation*}x=-\frac{b}{2a}\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}

Which is the quadratic equation formula.

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November 26, 2024 · 7:07 am

Infinite Product Expansion of cos (x)

Remember

(1)   \begin{equation*}cos(x)=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation*}

We know that cos(x)=0 for odd integer multiples of \frac{\pi}{2}, i.e. \frac{\pi}{2}, \frac{3\pi}{2}, ..., which is \frac{(2n-1)\pi}{2} for n\neq 0

Hence,

    \begin{equation*}0=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation}

for x=\frac{(2n-1)\pi}{2}, n>0

We can factorise our cos(x) expansion

    \begin{equation*}(1-\frac{x^2}{r_1})(1-\frac{x^2}{r_2})...\end{equation}

We know r_1=\frac{\pi}{2}, r_2=\frac{3\pi}{2}, ...

    \begin{equation*}cos(x)=(1-\frac{x^2}{(\frac{\pi}{2})^2})(1-\frac{x^2}{(\frac{3\pi}{2})^2})...(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}

    \begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}

    \begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{4x^2}{(2n-1)^2\pi^2})\end{equation}

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September 19, 2024 · 9:43 am

Solving Cubic Functions

I have been thinking about cubics a bit lately because some of my students are solving and then sketching cubics. Plus I am reading An Imaginary Tale by Paul Nahin, which talks about solving cubics and complex numbers.

Cubics must have at least one real root. If one of the roots is a rational number, then we can use the Factor and Remainder Theorem.

For example,

Solve 2x^3-3x^2-3x+2=0

We know the root(s) must be a factor of 2\times3=6.
I always start with 1 or -1
2(1)^3-3(1)^2-3(1)+2=2-3-3+2=-2 \therefore x\neq=1
Try -1
2(-1)^3-3(-1)^2-3(-1)+2=-2-3+3+2=0 \therefore x=-1 and (x+1) is a factor.
Then we can do polynomial long division.

Now we know that 2x^3-3x^2-3x+2=(x+1)(2x^2-5x+2)
And we can factorise the quadratic (or using the quadratic equation formula)
2x^2-5x+2=2x^2-4x-x+2
=2x(x-2)-1(x-2)
=(2x-1)(x-2)
\therefore x=-1, \frac{1}{2}, 2

But what if it is not factorisable?

For example,

Solve 2x^3+5x^2-2x+4=0

How many roots does this equation have?

We could find the derivative and find out how many stationary points the function has.

f'(x)=6x^2+10x-2

This is a quadratic function. Find the discriminant to determine the number of roots.

\Delta=b^2-4ac=100-4(6)(-2)=148

As \Delta>0, there are two stationary points, which means we could have 1, 2 (one root is repeated) or three roots, depending on if the function crosses the x-axis between stationary points. So not much use.

We could try the discriminant of a cubic.

\Delta=18abcd-4b^3d+b^2c^2-4ac^2-27a^2d^2

\Delta=18(2)(5)(-2)(4)-4(5^3)(4)+(5^2)(-2)^2-4(2)(-2)^2-27(2)^2(4)^2=-5100

The discriminant is negative so there is one real root.

From my reading, we need to turn the cubic into a depressed cubic (cubics of the form x^3+px+q=0).

We can do this by using a change of variable.

Let x=t-\frac{5}{6}
2(x^3+\frac{5}{2}x^2-x+2)=0
\therefore x^3+\frac{5}{2}x^2-x+2=0
Substitute t-\frac{5}{6} into the cubic.
(t-\frac{5}{6})^3+\frac{5}{2}(t-\frac{5}{6})^2-(t-\frac{5}{6})+2
(t^3-3(\frac{5}{6})t^2+3(\frac{5}{6})^2t-(\frac{5}{6})^3+\frac{5}{2}(t^2-2(\frac{5}{6})t+(\frac{5}{6})^2)-t+\frac{5}{6}+2
t^3-\frac{5}{2}t^2+\frac{25}{12}t-\frac{125}{216}+\frac{5}{2}t^2-\frac{25}{6}t+\frac{125}{72}-t+\frac{5}{6}+2
t^3-\frac{37}{12}t+\frac{431}{108}

We can then use Cardano’s formula

x=\sqrt[3]{-\frac{q}{2}+\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}

p=-\frac{37}{12} and q=\frac{431}{108}
\frac{p}{3}=\frac{-37}{36}
\frac{q}{2}=\frac{431}{216}
(\frac{431}{216})^2+(\frac{-37}{36})^3=\frac{139}{48}
t=\sqrt[3]{\frac{-431}{216}+\sqrt{\frac{139}{48}}}+\sqrt[3]{\frac{-431}{216}-\sqrt{\frac{139}{48}}}
t=-2.21095...
\therefore x=-2.21095...-\frac{5}{6}=-3.044

We can see from the sketch below that there is only one solution and it is about -3.

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September 14, 2024 · 9:45 am

Sum and Product of the Roots of Polynomials

There is a relationship between the sum and product of the roots of a polynomial and the co-efficient of the polynomial.

Let’s start with a quadratic.

The general form for a quadratic (polynomial of degree 2) is

y=ax^2+bx+c

Use the quadratic equation formula to find the roots

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Hence the roots are

x_1=\frac{-b+\sqrt{b^2-4ac}}{2a} and x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}

Sum of the roots:

\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-2b}{2a}=\frac{-b}{a}

Product of the roots:

\frac{-b+\sqrt{b^2-4ac}}{2a}\times\frac{-b-\sqrt{b^2-4ac}}{2a}

\frac{b^2}{4a^2}-\frac{b^2-4ac}{4a^2}

\frac{4ac}{4a^2}

\frac{a}{c}

Worked Example
The equation 4x^2+bx+c=0 has two distinct roots. The product of the roots is \frac{3}{4} and the sum is 2. Find b and c.
\frac{-b}{a}=2
a=4
\frac{-b}{4}=2
b=-8

\frac{c}{a}=\frac{3}{4}
\frac{c}{4}=\frac{3}{3}
c=3
The equations is 4x^2-8x+3

Solve the equation to prove the roots do in fact sum to 2 and multiply to \frac{3}{4}
4x^2-8x+3
4x^2-6x-2x+3
2x(2x-3)-1(2x-3)
(2x-1)(2x-3)
x_1=\frac{1}{2} and x_2=\frac{3}{2}
\frac{1}{2}+\frac{3}{2}=2 and \frac{1}{2}\times\frac{3}{2}=\frac{3}{4}

Let’s move to a cubic function.

The general equation is f(x)=ax^3+bx^2+cx+d

Let’s say the roots of this cubic are \alpha, \beta, \gamma

Then ax^3+bx^2+cx+d=a(x-\alpha)(x-\beta)(x-\gamma)

=a(x^2-\beta x - \alpha x+\alpha\beta)(x-\gamma)

=a(x^2-(\alpha+\beta)x+\alpha\beta)(x-\gamma)

=a(x^3-\gamma x^2-x^2(\alpha+\beta)+\gamma(\alpha+\beta)x-\alpha\beta\gamma)

=a(x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma)

The sum of the roots

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

The product of the roots

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

Also, it can be handy to know

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

Worked example
f(x)=x^3-6x^2+4x+12, the roots are \alpha, \beta and \gamma
Find
(a) \alpha+\beta+\gamma
(b) \alpha^2+\beta^2+\gamma^2
(c) \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}

(a) \alpha+\beta+\gamma=\frac{-b}{a}
\alpha+\beta+\gamma=\frac{6}{1}
\alpha+\beta+\gamma=6

(b)\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2\alpha\beta-2\alpha\gamma-2\beta\gamma
=6^2-2(4)
=28

(c)\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta\gamma}{\alpha\beta\gamma}+\frac{\alpha\gamma}{\alpha\beta\gamma}+\frac{\alpha\beta}{\alpha\beta\gamma}
=\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}
=\frac{4}{-13}
=\frac{-1}{3}

We can extend the method we used for finding the sum and product of the roots of cubic to polynomials of greater degree.

If the four roots of a quartic are \alpha, \beta, \gamma and \delta, and the general equation is ax^4+bx^3+cx^2+dx+e, then

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

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

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

\alpha\beta\gamma\delta=\frac{e}{a}

Worked Example (just one more)
The roots of the cubic equation x^3-4x^2-3x-2 are \alpha, \beta and \gamma. Find the cubic equation whose roots are \alpha+\beta, \alpha+\gamma, and \beta+\gamma

\alpha+\beta+\gamma=4
\alpha\beta+\alpha\gamma+\beta\gamma=-3
\alpha\beta\gamma=2

\frac{-b}{a}=\alpha+\beta+\alpha+\gamma+\beta+\gamma
\frac{-b}{a}=2(\alpha+\beta+\gamma
\frac{-b}{a}=2(4)
\frac{-b}{a}=8

\frac{c}{a}=(\alpha+\beta)(\alpha+\gamma)+(\alpha+\beta)(\beta+\gamma)+(\alpha+\gamma)(\beta+\gamma)
=\alpha^2+\alpha\gamma+\beta\gamma+\gamma^2+\alpha\beta+\alpha\gamma+\beta^2+\beta\gamma+\alpha\beta+\alpha\gamma+\gamma\beta+\gamma^2
=\alpha^2+\beta^2+\gamma^2+3\alpha\gamma+3\beta\gamma+3\alpha\beta
=(\alpha+\beta+\gamma)^2-2\alpha\gamma-2\beta\gamma-2\alpha\beta+3\alpha\gamma+3\beta\gamma+3\alpha\beta
=(\alpha+\beta+\gamma)^2+\alpha\gamma+\alpha\beta+\beta\gamma
=4^2-3
\frac{c}{a}=13

\frac{-d}{a}=(\alpha+\beta)(\alpha+\gamma)(\beta+\gamma)
=(\alpha^2+\alpha\gamma+\beta\alpha+\beta\gamma)(\beta+\gamma)
=\alpha^2\beta+\alpha^2\gamma+\alpha\gamma\beta+\alpha\gamma^2+\beta^2\alpha+\beta\alpha\gamma+\beta^2\gamma+\beta\gamma^2
=2\alpha\beta\gamma+\alpha^2\beta+\beta^2\alpha+\alpha^2\gamma+\gamma^2\alpha+\beta^2\gamma+\gamma^2\beta
=2(2)+\alpha\beta(\alpha+\beta)+\alpha\gamma(\alpha+\gamma)+\beta\gamma(\beta+\gamma)
=-24+\alpha\beta(\alpha+\beta+\gamma)-\alpha\beta\gamma+\alpha\gamma(\alpha+\gamma+\beta)-\alpha\beta\gamma+\beta\gamma(\beta+\gamma+\alpha)-\beta\gamma\alpha
=4+(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)-3(2)
=-2+(4)(-3)
\frac{-d}{a}=-14

If a=1 then b=-8, c=13 and d=14
The cubic is x^3-8x^2+13x+14

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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
7^01
7^17
7^249
7^3343
7^42401
7^516807
As numbersAs Powers of 7
3438=1\times2401+10373438=1\times7^4+1037
1037=3\times343+81037=3\times7^3+8
8=1\times7+18=1\times7^1+1
1=1\times11=1\times7^0

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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