# Generalisation of $\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right)$

Hello every $\frac { 45\pi }{ 4096 } \left( \begin{matrix} 7 \\ 2.5 \end{matrix} \right)$.

I hope you are having a great time!

I have presumably found a way to solve

$\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right)$

if ${ r }_{ n }$ are the roots to any equation of the form $f\left( x \right)={ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }$

and $m$ is any real or complex number.

Claim:

$\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }$

if $f\left( x \right)={ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }$

Simply put ${ a }_{ n }$ is the coefficient of ${ x }^{ n }$ in $f\left( x \right)$ or even simpler,

${ a }_{ n }$ is the leading coefficient of $f\left( x \right)$.

Example

Lets have an explicit example where $m=3$ and $f\left( x \right) =2{ x }^{ 3 }+12{ x }^{ 2 }-14x-120$

So $\left( { r }_{ 1 }+3 \right) \left( { r }_{ 2 }+3 \right) \left( { r }_{ 3 }+3 \right) ={ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27$

Then using Vieta's, ${ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27=-\frac { -120 }{ 2 } +3\times \frac { -14 }{ 2 } -9\times \frac { 12 }{ 2 } +27\\ =12$

and indeed ${ \left( -1 \right) }^{ 3 }\frac { f\left( -3 \right) }{ 2 } =12$

and also the roots of $f\left( x \right) =2{ x }^{ 3 }+12{ x }^{ 2 }-14x-120$ is $-5$,$-4$ and $3$. Meaning the product is $\left( -5+3 \right) \left( -4+3 \right) \left( 3+3 \right) =12$

Proof 1

$\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ m }^{ 0 }\prod _{ y=1 }^{ n }{ { r }_{ n } } +{ m }^{ 1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 }{ r }_{ 2 }\dots { r }_{ n-1 } } \right) +\cdots +{ m }^{ n-1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 } } \right) +{ m }^{ n }\left( \sum _{ n=1 }^{ n }{ 1 } \right)$

Using Vieta's, ${ m }^{ 0 }\prod _{ y=1 }^{ n }{ { r }_{ y } } +{ m }^{ 1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 }{ r }_{ 2 }\dots { r }_{ n-1 } } \right) +\cdots +{ m }^{ n-1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 } } \right) +{ m }^{ n }\left( \sum _{ n=1 }^{ n }{ 1 } \right) ={ \left( -1 \right) }^{ n }\frac { { m }^{ 0 }{ a }_{ 0 } }{ { a }_{ n } } +\cdots { \left( -1 \right) }^{ 1 }\frac { { m }^{ n-1 }{ a }_{ n-1 } }{ { a }_{ n } } +{ \left( -1 \right) }^{ 0 }\frac { { m }^{ n }{ a }_{ n } }{ { a }_{ n } } \\ =\sum _{ y=0 }^{ n }{ \frac { { \left( -1 \right) }^{ y }{ m }^{ n-y }{ a }_{ n-y } }{ { a }_{ n } } } \\ ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }$

Proof 2

With the basic definition of a polynomial,

${ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }={ a }_{ n }\left( x-{ r }_{ 1 } \right) \left( x-{ r }_{ 2 } \right) \cdots \left( x-{ r }_{ n } \right)$

$\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ \left( -1 \right) }^{ n }\left( -m-{ r }_{ 1 } \right) \left( -m-{ r }_{ 2 } \right) \cdots \left( -m-{ r }_{ n } \right) \\ ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }$

$\left( k{ r }_{ 1 }+m \right) \left( k{ r }_{ 2 }+m \right) \dots \left( { kr }_{ n }+m \right)$

for $k$ being any real or complex number.

We too also can solve this by take $k$ out in each term and then use the generalisation to solve

$\left( k{ r }_{ 1 }+m \right) \left( k{ r }_{ 2 }+m \right) \dots \left( { kr }_{ n }+m \right) ={ k }^{ n }\left( { r }_{ 1 }+\frac { m }{ k } \right) \left( { r }_{ 2 }+\frac { m }{ k } \right) \cdots \left( { r }_{ n }+\frac { m }{ k } \right) \\ ={ \left( -1 \right) }^{ n }{ k }^{ n }\frac { f\left( -\frac { m }{ k } \right) }{ { a }_{ n } } \\ ={ \left( -k \right) }^{ n }\frac { f\left( -\frac { m }{ k } \right) }{ { a }_{ n } }$

Phew...

I really do hope this note helped you in one way or another! Well, that's probably it! Note by Joel Yip
3 years, 10 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Nice note! @Joel Yip

Yes, I think this is a good generalization for such sums. No wonder we'll get to see more good problems from many others now :)

- 3 years, 10 months ago

thanks!

- 3 years, 10 months ago

Perhaps you can add questions into this!

- 3 years, 10 months ago

- 3 years, 10 months ago

great note @Joel Yip !!!!!

- 3 years, 10 months ago

thank you!

- 3 years, 10 months ago

Should I give more examples?

- 3 years, 10 months ago

nice work Joel. thumbs up!!

- 3 years, 10 months ago

Thank you to all of the positive replies to this note! I just want to say thanks again!

- 3 years, 10 months ago

Good Note. Nice work Joel Yip.

- 3 years, 9 months ago