Serious Polynomial Fun

**nath_quam** · June 12th 2006, 11:46 PM

This is for anyone who wants a challenge with polynomials because they are too difficult for me thank you for any help

i have found the answers

a)-2
b)-11
c)12
d)-11/12
e)-1/6
f)0
g)-132
h)26
i)13/72

would someone be able to check these

**earboth** · June 13th 2006, 06:48 AM

Originally Posted by nath_quam

i have found the answers for
a)-2
b)-11
c)12
d)-11/12
e)-1/6
f)0
g)?
h)26
i)?
would someone be able to check these and solve/tell me how to do the unknowns

Hello,

all your results are correct. (Congratulations!)

with g): Plug in the values you know and you'll have:
$(-12)^2\cdot (-1)+(-3)^2\cdot (-4)+(4)^2\cdot 3 = -132$

with i): You may know that $n^{-2}=\frac{1}{n^2}$ . With your solutions you get:

$\left( {1\over(-12)}\right)^2+\left( {1\over(-3)}\right)^2+\left( {1\over(4)}\right)^2=\left( {13\over 72}\right)$

Greetings

EB

**Soroban** · June 13th 2006, 02:04 PM

Hello, nath_quam!

These are great problems!
They give me a chance to show-off . . . LOL!

The first three are easy . . . if you know this theorem:
If $a,b,c$ are roots of the cubic $x^3 + px^2 + qx + r\,=\,0$
. . then: . $a + b + c \,= \,-p,\;\;ab + bc + ac \,= \,q,\;\;pqr \,= \,-r$

If $a,b,c$ are roots of $x^3 + 2x^2 - 11x - 12\:=\:0$ , find:

$a)\;\;a + b + c \;=\;-2$
$b)\;\;ab + ac + bc\;=\;-11$
$c)\;\;abc\;=\;12$

Then it gets interesting!
These problems lend themselves to some clever manipulations.

$d)\;\;\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\;=\;\frac{ab + ac + bc}{abc} \;= \;\frac{-11}{12}\;=\;-\frac{11}{12}$

$e)\;\;\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\;=\;\frac{a + b + c}{abc}\;=\;\frac{-2}{12}\;=\;-\frac{1}{6}$

$f)\;\;(a + 1)(b + 1)(c + 1)\;=\;abc + (ab + ac + bc) +$ $(a + b + c) + 1$
. . . . . $=\;12 + (-11) + (-2) + 1 \;= \;0$

$g)\;\;(ab)^2c + (ac)^2b + (bc)^2a \;=\;a^2b^2c + a^2bc^2 + ab^2c^2$
. . . . . $= \;abc(ab + ac + bc) \;= \;(12)(-11) \;= \; -132$

$h)\;\;a^2 + b^2 + c^2$ . . . This one requires Olympic-level gymnastics.

We know that: . $a + b + c = -2$
. . . . Hence: . $(a + b + c)^2 \;=\;4$

Expand: . $a^2 + 2ab + 2ac + b^2 + 2bc + c^2 \;=\;4$

We have: . $(a^2 + b^2 + c^2) + 2(ab + bc + ac) \;=\;4$

Hence: . $(a^2 + b^2 + c^2) + 2(-11)\;=\;4\quad\Rightarrow\quad a^2+b^2+c^2\:=\:26$

$i)\;\;(ab)^{-2} + (ac)^{-2} + (bc)^{-2}\;=\;\frac{1}{a^2b^2} + \frac{1}{a^2c^2} +$ $\frac{1}{b^2c^2} \;= \;\frac{a^2 + b^2 + c^2}{a^2b^2c^2}$

. . . . . $=\;\frac{a^2 + b^2 + c^2}{(abc)^2}\;=\;\frac{26}{12^2}\;=\;\frac{13}{72 }$

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