Polynomial coefficient problem

**PaulAdrienMauriceDirac** · October 31st 2012, 05:16 AM

Hello, I have an another unsolved problem from Gelfand's book on Algebra, here it is:
Problem 153. The polynomial $P=x^3+x^2-10x+1$ has three different roots (the authors guarantee it) denoted by $x_1,x_2,x_3$ . Write a polynomial with integer coefficients having roots
(a) $x_1+1, x_2+1, x_3+1$ ; (b) $2x_1, 2x_2, 2x_3$ ; (c) $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}$

I understand that $P=x^3+x^2-10x+1=(x-x_1)(x-x_2)(x-x_3)$ , but that's pretty much all I can 'do', hints would be appreciated, thanks in advance.

**Soroban** · October 31st 2012, 06:38 AM

Hello, PaulAdrienMauriceDirac!

I'll do part (a).
I hope you understand the reasoning.

$\text{Polynomial }P(x)\:=\:x^3+x^2-10x+1\,\text{ has three roots, denoted }a,\,b,\,c.$

$\text{Write a polynomial }Q(x)\text{ with integer coefficients having roots:}$

$(a)\;a+1,\:b+1,\:c+1 \qquad (b)\;2a,\,2b,\,2c \qquad (c)\;\frac{1}{a},\:\frac{1}{b},\:\frac{1}{c}$

We know that: . $\begin{Bmatrix} a+b+c \:=\:\text{-}1 \\ ab+bc+ac \:=\:\text{-}10 \\ abc \:=\:\text{-}1 \end{Bmatrix}$

$(a)\text{ Roots: }\:a+1,\:b+1,\:c+1$

Sum of roots: . $(a+1)+(b+1)+(c+1) \;=\;(a+b+c) + 3 \;=\;\text{-}1 + 3 \;=\;2$

Sum of pairs: . $(a+1)(b+1) + (b+1)(c+1) + (a+1)(c+1)$

. . . . . . . . . $=\;(ab + bc + ac) + 2(a+b+c) + 1$

. . . . . . . . . $=\;(\text{-}10) + 2(\text{-}1) + 3\;=\;\text{-}9$

Product of roots: . $(a+1)(b+1)(c+1)$

. . . . . . . . . . . $=\;abc + (ab + bc + ac) + (a+b+c) + 1$

. . . . . . . . . . . $=\;\text{-}1 + (\text{-}10) + (\text{-}1) + 1 \;=\;\text{-}11$

Therefore: . $Q(x) \;=\;x^3 - 2x^2 - 9x + 11$

**PaulAdrienMauriceDirac** · October 31st 2012, 07:17 AM

Originally Posted by Soroban

Hello, PaulAdrienMauriceDirac!

I'll do part (a).
I hope you understand the reasoning.

We know that: . $\begin{Bmatrix} a+b+c \:=\:\text{-}1 \\ ab+bc+ac \:=\:\text{-}10 \\ abc \:=\:\text{-}1 \end{Bmatrix}$

$(a)\text{ Roots: }\:a+1,\:b+1,\:c+1$

Sum of roots: . $(a+1)+(b+1)+(c+1) \;=\;(a+b+c) + 3 \;=\;\text{-}1 + 3 \;=\;2$

Sum of pairs: . $(a+1)(b+1) + (b+1)(c+1) + (a+1)(c+1)$

. . . . . . . . . $=\;(ab + bc + ac) + 2(a+b+c) + 1$

. . . . . . . . . $=\;(\text{-}10) + 2(\text{-}1) + 3\;=\;\text{-}9$

Product of roots: . $(a+1)(b+1)(c+1)$

. . . . . . . . . . . $=\;abc + (ab + bc + ac) + (a+b+c) + 1$

. . . . . . . . . . . $=\;\text{-}1 + (\text{-}10) + (\text{-}1) + 1 \;=\;\text{-}11$

Therefore: . $Q(x) \;=\;x^3 - 2x^2 - 9x + 11$

Thank you very much! I didn't know that Vieta's theorem existed for cubic equations, your explanation is very clear.

**Salahuddin559** · October 31st 2012, 08:35 PM

Hi PaulAdrienMauriceDirac,

P(x) = (x - x1)(x - x2)(x - x3) = x^3 - (x1 + x2 + x3)x^2 + (x1x2 + x1x3 + x2x3)x - x1x2x3.
So, equating co-efficients like this holds for cubics, and in general, all polynomial equations roots.

Salahuddin
Maths online

**MarkFL** · October 31st 2012, 08:55 PM

While Vieta's theorem is the way to go for parts b) and c), for part a) you could simply shift the polynomial 1 unit to the right to add 1 to the roots:

$f(x)=(x-1)^3+(x-1)^2-10(x-1)+1=$

$x^3-3x^2+3x-1+x^2-2x+1-10x+10+1=$

$x^3-2x^2-9x+11$

**PaulAdrienMauriceDirac** · October 31st 2012, 09:43 PM

Originally Posted by MarkFL2

While Vieta's theorem is the way to go for parts b) and c), for part a) you could simply shift the polynomial 1 unit to the right to add 1 to the roots:

$f(x)=(x-1)^3+(x-1)^2-10(x-1)+1=$

$x^3-3x^2+3x-1+x^2-2x+1-10x+10+1=$

$x^3-2x^2-9x+11$

Oh, that's so much easier, too bad I haven't thought about it, thank you very much!

**BobP** · November 1st 2012, 02:17 AM

The same sort of method will work for part (b) as well, replace x by x/2.
The resulting polynomial is
$\frac{x^{3}}{8}+\frac{x^{2}}{4}-5x+1.$

**PaulAdrienMauriceDirac** · November 1st 2012, 02:58 AM

Originally Posted by BobP

The same sort of method will work for part (b) as well, replace x by x/2.
The resulting polynomial is
$\frac{x^{3}}{8}+\frac{x^{2}}{4}-5x+1.$

Hm, are you sure that that method works on part (b) as well? because I got different results when using Vieta's theorem.

$P(x)=x^3+px^2+qx+r$
We know from Vieta's that:
$x_1+x_2+x_3=-p;$

$x_1x_2+x_1x_3+x_2x_3=q;$

$x_1x_2x_3=-r$

So:
$x_1+x_2+x_3=-1$ -> $2x_1+2x_2+2x_3=2(x_1+x_2+x_3)=2(-1)=-2$
$x_1x_2+x_1x_3+x_2x_3=-10$ -> $4x_1x_2+4x_1x_3+4x_2x_3=4(x_1x_2+x_1x_3+x_2x_3)=-40$
$x_1x_2x_3=-1$ -> $2x_1*2x_2*2x_3=8(x_1x_2x_3)=8(-1)=-8$

Therefore:
$p=2, q=-40, r=8$ and $Q(x)=x^3+2x^2-40x+8$

**BobP** · November 1st 2012, 03:56 AM

Your polynomial is simply eight times mine, they will have the same zeros.

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