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(m - a)(m - b)(m - c) . . . (m - z) =

Guest May 29, 2015

Best Answer 

 #3
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+13

It contains a factor (m - m) which is equal to zero, so the answer is zero.

Guest May 30, 2015
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18+0 Answers

 #1
avatar+78575 
+13

 

Notice the following patterns:

 

(m -a)(m-b) = [m^2 - ma - mb + (ab)] = [ m^2 - (a+b)m + (ab)]

 

[ m^2 - (a+b)m + (ab)] [m - c ]  =

 

m^3 - (a+b)m^2 + (ab)m - cm^2 + (ac)m + (bc)m - abc =

 

m^3 - (a + b + c)m^2 + (ab + ac + bc)m - abc

 

Let (a,b,c,d,e.......z)  be denoted as SetZ.....so we have.....

 

(m-a)(m-b)(m-c)......(m-z)  =

 

m^26 - (sum of all elements in SetZ taken one at a time)m^25 + (sum of all elements in SetZ taken two at a time)m^24 - (sum of all elements of SetZ taken three at a time)m^23 +.........+ (sum of all elements in SetZ taken 24 at a time)m^2 -(sum of all elements of SetZ taken 25 at a time)m + (the product of all elements in SetZ)

 

I believe this is correct......!!!  {Can someone else - Melody, Alan?? - check my logic}

 

 

CPhill  May 29, 2015
 #2
avatar+26322 
+5

Yes, you're right Chris.

.

Alan  May 30, 2015
 #3
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+13
Best Answer

It contains a factor (m - m) which is equal to zero, so the answer is zero.

Guest May 30, 2015
 #4
avatar+78575 
+5

Yep, Anonymous.....I'm afraid you are correct....I had forgotten about that small detail!!!

 

 

CPhill  May 30, 2015
 #5
avatar+90988 
+5

Great thinking anon, are you the same anon that posted the question ??

 

I love this trick question  !!!

 

I just added it into the puzzles sticky thread :))

Melody  May 30, 2015
 #6
avatar+26322 
+5

We should note that Chris's expansion is still correct.  It's just that the resulting grand sum evaluates to zero!

.

Alan  May 30, 2015
 #7
avatar+90988 
0

YES absolutely  !!

Melody  May 30, 2015
 #8
avatar+78575 
0

Yep, Alan....it's just that ....if I had seen that up front.....I could have had 15 more minutes in my life devoted to some other pursuit.....LOL!!!!

 

 

 

 

CPhill  May 30, 2015
 #9
avatar+78575 
0

Melody......what are you gonna' title this one in the Puzzles Thread???....."Another One That Chris Screwed Up "   ????....... LOL!!!!!

 

CPhill  May 30, 2015
 #10
avatar+90988 
+5

Yes you could have been designing the next genaration of Marsian rockets.  

OR

Cleaning out your sock drawer.     LOL

Melody  May 30, 2015
 #11
avatar+90988 
0

As Alan pointed out, you did not s***w it up!  If I was here I would not  have even attempted it.

I would have  been doing something more relevant to me like looking for your odd socks.  Do you think that they could have made it to Australia.  There have been some big tides lately :/

Melody  May 30, 2015
 #12
avatar+78575 
0

My "odd socks" could have made it to Australia......they've been known to "walk on water"

 

 

CPhill  May 30, 2015
 #13
avatar+18712 
+5
heureka  May 30, 2015
 #14
avatar+1035 
+5

Classify it under Rube Goldberg Solutions for complex math problems with trivial non-roman ZEROS.

Sub-classification: Put a Sock in it.

Nauseated  May 30, 2015
 #15
avatar+414 
+3

(m -a)(m-b) = [m^2 - ma - mb + (ab)] = [ m^2 - (a+b)m + (ab)] [ m^2 - (a+b)m + (ab)] [m - c ] = m^3 - (a+b)m^2 + (ab)m - cm^2 + (ac)m + (bc)m - abc = m^3 - (a + b + c)m^2 + (ab + ac + bc)m - abc Let (a,b,c,d,e.......z) be denoted as SetZ.....so we have..... (m-a)(m-b)(m-c)......(m-z) = m^26 - (sum of all elements in SetZ taken one at a time)m^25 + (sum of all elements in SetZ taken two at a time)m^24 - (sum of all elements of SetZ taken three at a time)m^23 +.........+ (sum of all elements in SetZ taken 24 at a time)m^2 -(sum of all elements of SetZ taken 25 at a time)m + (the product of all elements in SetZ)

MaiaMitchell  Aug 9, 2015
 #16
avatar+414 
+3

Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic $x^2+ax+b=0$ with solutions $p$ and $q$, then we know that we can factor it as $x^2+ax+b=(x-p)(x-q)$ (Note that the first term is $x^2$, not $ax^2$.) Using the distributive property to expand the right side we get $x^2+ax+b=x^2-(p+q)x+pq$ We know that two polynomials are equal if and only if their coefficients are equal, so $x^2+ax+b=x^2-(p+q)x+pq$ means that $a=-(p+q)$ and $b=pq$. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the $x$ term. A similar set of relations for cubics can be found by expanding $x^3+ax^2+bx+c=(x-p)(x-q)(x-r)$. We can state Vieta's formula's more rigorously and generally. Let $P(x)$ be a polynomial of degree $n$, so $P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0$, where the coefficient of $x^{i}$ is ${a}_i$ and $a_n \neq 0$. As a consequence of the Fundamental Theorem of Algebra, we can also write $P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$, where ${r}_i$ are the roots of $P(x)$. We thus have that $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).$ Expanding out the right hand side gives us $a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.$ The coefficient of $x^k$ in this expression will be the $k$th symmetric sum of the $r_i$. We now have two different expressions for $P(x)$. These must be equal. However, the only way for two polynomials to be equal for all values of $x$ is for each of their corresponding coefficients to be equal. So, starting with the coefficient of $x^n$, we see that $a_n = a_n$ $a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$ $a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$ $\vdots$ $a_0 = (-1)^n a_n r_1r_2\cdots r_n$ More commonly, these are written with the roots on one side and the $a_i$ on the other (this can be arrived at by dividing both sides of all the equations by $a_n$). If we denote $\sigma_k$ as the $k$th symmetric sum, then we can write those formulas more compactly as $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for $1\le k\le {n}$. Problems Beginner Let $r_1,r_2,$ and $r_3$ be the three roots of the cubic $x^3+3x^2+4x-4$. Find the value of $r_1r_2+r_1r_3+r_2r_3$. Suppose the polynomial $5x^3+4x^2-8x+6$ has three real roots $a,b$, and $c$. Find the value of $a(1+b+c)+b(1+a+c)+c(1+a+b)$. Let $m$ and $n$ be the roots of the quadratic equation $4x^2 + 5x + 3 = 0$. Find $(m + 7)(n + 7)$. Intermediate Let $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$. Find $a,b,c$ And The $k$-th elementary symmetric sum of a set of $n$ numbers is the sum of all products of $k$ of those numbers ($1 \leq k \leq n$). For example, if $n = 4$, and our set of numbers is $\{a, b, c, d\}$, then: 1st Symmetric Sum = $S_1 = a+b+c+d$ 2nd Symmetric Sum = $S_2 = ab+ac+ad+bc+bd+cd$ 3rd Symmetric Sum = $S_3 = abc+abd+acd+bcd$ 4th Symmetric Sum = $S_4 = abcd$ Notation The first elementary symmetric sum of $f(x)$ is often written $\sum_{sym}f(x)$. The $n$th can be written $\sum_{sym}^{n}f(x)$ Uses Any symmetric sum can be written as a polynomial of the elementary symmetric sum functions. For example, $x^3 + y^3 + z^3 = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - xz) + 3xyz = S_1^3 - 3S_1S_2 + 3S_3$. This is often used to solve systems of equations involving power sums, combined with Vieta's formulas. Elementary symmetric sums show up in Vieta's formulas. In a monic polynomial, the coefficient of the $x^1$ term is $e_1$, and the coefficient of the $x^k$ term is $e_k$, where the symmetric sums are taken over the roots of the polynomial.(done finally)

MaiaMitchell  Aug 9, 2015
 #17
avatar
0

Wow! Did you hurt your finger pasting all that in there?

Guest Aug 9, 2015
 #18
avatar+414 
0

Half of it was pasted half of it was written anyonamous 

MaiaMitchell  Aug 10, 2015

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