# Thread: Polynomials question

1. ## Polynomials question

For roots of a polynomial stuff, do you have to remember like alpha^2 + beta^2 + gamma^2 + delta^2 and what this is equivalent to? I know alpha^2 + beta^2 = (alpha+beta)^2 - 2alpha*beta but do we have to remember the hard ones like shown above? If we do is there a way to derive them?

2. ## Re: Polynomials question

Well you could form a polynomial with roots alpha^2, beta^2, gamma^2, delta^2. And then finding the sum of roots (-b/a) of that polymomial

3. ## Re: Polynomials question

Originally Posted by pikachu975
For roots of a polynomial stuff, do you have to remember like alpha^2 + beta^2 + gamma^2 + delta^2 and what this is equivalent to? I know alpha^2 + beta^2 = (alpha+beta)^2 - 2alpha*beta but do we have to remember the hard ones like shown above? If we do is there a way to derive them?
$\noindent It's worth knowing that a^2 + b^2 + c^2 + d^2 = \left(a+b+c+d \right)^{2} -2\left(ab + bc + cd + da + ac + bd \right). In general,$

\begin{align*}\sum_{i = 1}^{n} a_{i}^{2} &= \left(\sum_{i=1}^{n} a_{i}\right)^{2}-2\sum_{1\leq i < j\leq n}a_{i}a_{j}.\end{align*}

$\noindent This just comes from rearranging the formula for the square of a sum. Note the last sum has \binom{n}{2} terms (it's all the two-way combinations of the a_{i}). In words, the formula is sum of squares = square of sum - twice the sum of pairwise products''.$

4. ## Re: Polynomials question

Originally Posted by InteGrand
$\noindent It's worth knowing that a^2 + b^2 + c^2 + d^2 = \left(a+b+c+d \right)^{2} -2\left(ab + bc + cd + da + ac + bd \right). In general,$

\begin{align*}\sum_{i = 1}^{n} a_{i}^{2} &= \left(\sum_{i=1}^{n} a_{i}\right)^{2}-2\sum_{1\leq i < j\leq n}a_{i}a_{j}.\end{align*}

$\noindent This just comes from rearranging the formula for the square of a sum. Note the last sum has \binom{n}{2} terms (it's all the two-way combinations of the a_{i}). In words, the formula is sum of squares = square of sum - twice the sum of pairwise products''.$
Ah thanks for that, didn't know this works for more than three terms

5. ## Re: Polynomials question

Originally Posted by InteGrand
$\noindent It's worth knowing that a^2 + b^2 + c^2 + d^2 = \left(a+b+c+d \right)^{2} -2\left(ab + bc + cd + da + ac + bd \right). In general,$

\begin{align*}\sum_{i = 1}^{n} a_{i}^{2} &= \left(\sum_{i=1}^{n} a_{i}\right)^{2}-2\sum_{1\leq i < j\leq n}a_{i}a_{j}.\end{align*}

$\noindent This just comes from rearranging the formula for the square of a sum. Note the last sum has \binom{n}{2} terms (it's all the two-way combinations of the a_{i}). In words, the formula is sum of squares = square of sum - twice the sum of pairwise products''.$
Is there a formula for the sum of cubes? Thanks for this

6. ## Re: Polynomials question

Originally Posted by pikachu975
Is there a formula for the sum of cubes? Thanks for this
yes

i derived it from scratch during my honours year... a fun exercise

7. ## Re: Polynomials question

Originally Posted by pikachu975
Is there a formula for the sum of cubes? Thanks for this
$\noindent Yes, there is. In fact, there is a recursive formula for all sums of non-negative integer powers (called \textbf{power sums}) in terms of lower powers' sums and elementary symmetric polynomials (expressions like sum of roots, sum of pairs of roots, etc., which are easily computable given a polynomial P(x)). But you're not expected to know them for the HSC. If you want to read up about them, you can do so here:$

https://en.wikipedia.org/wiki/Newton's_identities .

As background reading too, you may want to see:

- https://en.wikipedia.org/wiki/Symmetric_polynomial

- https://en.wikipedia.org/wiki/Elemen...ric_polynomial (and this part: https://en.wikipedia.org/wiki/Elemen...ic_polynomials (which was asked as a question by glittergal96 in a previous MX2 Advanced Level marathon here: HSC 2016 MX2 Marathon ADVANCED (archive))).

8. ## Re: Polynomials question

$\noindent You may want to keep in mind the three variable result$

$a^{3} + b^{3} + c^{3} - 3abc = \left(a + b + c\right)\left(a^{2} + b^{2} + c^{2} - ab - bc - ca\right).$

$\noindent This can also be used to prove three-variable AM-GM, by first showing (a^{2} + b^{2} + c^{2} \geq ab + bc + ca for all non-negative real numbers a,b,c with equality iff a = b = c.$

9. ## Re: Polynomials question

Is there any way to do all this stuff using pythagorus? I mean with all the a^2, b^2

10. ## Re: Polynomials question

Originally Posted by fluffy54324
Is there any way to do all this stuff using pythagorus? I mean with all the a^2, b^2
uhhh good pattern recognition, but pythagoras is for triangles/orthogonality only

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