Welcome to our community

Be a part of something great, join today!

Determine the product value

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,689
Hi MHB,

It's me again...I find this problem to be very interesting yet very difficult to me. I tried to approach it using the Vieta's formula, knowing the given function $p(x)$ has only one real root and 4 complex roots, where I let the 4 complex roots be $a\pm bi$ and $c\pm di$, but it failed me. I start to think this must be a problem that is not compatible to my level and thus I post it here hoping to find someone who will be interested with it and solve it for me...:eek:

Thanks in advance!(Sun)

Problem:

Let \(\displaystyle p(x)=x^5+x^2+1\) have roots $r_1$, $r_2$, $r_3$, $r_4$, and $r_5$. Let \(\displaystyle q(x)=x^2-2\). Determine the product of \(\displaystyle q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)\).
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Let \(\displaystyle p(x)=x^5+x^2+1\) have roots $r_1$, $r_2$, $r_3$, $r_4$, and $r_5$. Let \(\displaystyle q(x)=x^2-2\). Determine the product of \(\displaystyle q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)\).
The product \(\displaystyle q(r_1)q(r_2)q(r_3)q(r_4)q(r_5)\) is a symmetric polynomial in $r_1,\dots,r_5$, and every symmetric polynomial can be expressed through elementary symmetric polynomials, which, in turn, can be expressed through the coefficients of $p(x)$ by the Vieta's formulas. Finding this expression is left as an exercise. (Smile)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Let \(\displaystyle p(x)=x^5+x^2+1\) have roots $r_1$, $r_2$, $r_3$, $r_4$, and $r_5$. Let \(\displaystyle q(x)=x^2-2\). Determine the product of \(\displaystyle q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)\).
If $y = x^2 - 2$ then $x = (y+2)^{1/2}$, and $x^5+x^2+1 = (y+2)^{5/2} + y+3$. So if $r_1,\ldots,r_5$ are the roots of $p(x)$ then $q(r_1),\ldots,q(r_5)$ are the roots of $(y+2)^{5/2} + y+3$. But if $(y+2)^{5/2} + y+3 = 0$ then $(y+2)^5 = (y+3)^2$, so that $y^5 + \ldots + (2^5-3^2) = 0$. The product of the roots of that polynomial is the negative of the constant term, namely $-(32-9) = -23.$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hi MHB,

It's me again...I find this problem to be very interesting yet very difficult to me. I tried to approach it using the Vieta's formula, knowing the given function $p(x)$ has only one real root and 4 complex roots, where I let the 4 complex roots be $a\pm bi$ and $c\pm di$, but it failed me. I start to think this must be a problem that is not compatible to my level and thus I post it here hoping to find someone who will be interested with it and solve it for me...:eek:

Thanks in advance!(Sun)

Problem:

Let \(\displaystyle p(x)=x^5+x^2+1\) have roots $r_1$, $r_2$, $r_3$, $r_4$, and $r_5$. Let \(\displaystyle q(x)=x^2-2\). Determine the product of \(\displaystyle q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)\).
Is...


$\displaystyle p(x) = \prod_{i=1}^{5} (x-r_{i})\ (1)$

... and...


$\displaystyle q(x) = (x-\sqrt{2}) (x+\sqrt{2})\ (2)$

... so that...


$\displaystyle \prod_{i=1}^{5} q(r_{i}) = \prod_{i=1}^{5} (r_{i} - \sqrt{2}) \prod_{i=1}^{5} (r_{i} + \sqrt{2}) = - p(\sqrt{2})\ p(-\sqrt{2}) = (3 - 2^{\frac{5}{2}})\ (3 + 2^{\frac{5}{2}}) = 9 - 32 = -23\ (3)$

Kind regards

$\chi$ $\sigma$
 
Last edited:
  • Thread starter
  • Admin
  • #5

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,689
The product \(\displaystyle q(r_1)q(r_2)q(r_3)q(r_4)q(r_5)\) is a symmetric polynomial in $r_1,\dots,r_5$, and every symmetric polynomial can be expressed through elementary symmetric polynomials, which, in turn, can be expressed through the coefficients of $p(x)$ by the Vieta's formulas. Finding this expression is left as an exercise. (Smile)
If $y = x^2 - 2$ then $x = (y+2)^{1/2}$, and $x^5+x^2+1 = (y+2)^{5/2} + y+3$. So if $r_1,\ldots,r_5$ are the roots of $p(x)$ then $q(r_1),\ldots,q(r_5)$ are the roots of $(y+2)^{5/2} + y+3$. But if $(y+2)^{5/2} + y+3 = 0$ then $(y+2)^5 = (y+3)^2$, so that $y^5 + \ldots + (2^5-3^2) = 0$. The product of the roots of that polynomial is the negative of the constant term, namely $-(32-9) = -23.$
Is...


$\displaystyle p(x) = \prod_{i=1}^{5} (x-r_{i})\ (1)$

... and...


$\displaystyle q(x) = (x-\sqrt{2}) (x+\sqrt{2})\ (2)$

... so that...


$\displaystyle \prod_{i=1}^{5} q(r_{i}) = \prod_{i=1}^{5} (r_{i} - \sqrt{2}) \prod_{i=1}^{5} (r_{i} + \sqrt{2}) = - p(\sqrt{2})\ p(\sqrt{2}) = (3 - 2^{\frac{5}{2}})\ (3 + 2^{\frac{5}{2}}) = 9 - 32 = -23\ (3)$

Kind regards

$\chi$ $\sigma$
Thank you, Evgeny.Makarov, Opalg and chisigma for the reply!!!

I truly appreciate the three of you taking the time out to reply to this thread because I learned a great deal from the replies...

I love you guys!(Inlove)