282. Factorize of polynomials.
by Virgil Nicula, Jun 4, 2011, 3:01 AM
PP1 (1976 Tokyo University entrance exam). Find 
such that the polynomial 
can be
factorized into the product of two polynomials with integer coefficients, positive degree. Then for such
factorize 
.
. Our polynomial admits the root 
or 
, i.e.
.
. Now appear two situations :
and in this case our polynomial becomes 
.
.
PP2 (Uzbekistan NMO 2011). Find such that 
is equal to the product of two non-constant polynomials with integer coefficients.
. Our polynomial admits the root or , i.e. 
and in this case 
.
. Now appear two situations :
. For 
our polynomial becomes 
.
For 
our polynomial becomes 
.
PP3. Find
such that the polynomial equation 
has exactly one real root.
Proof 1. has exactly one real root 
has exactly one real root 
the graphs 
, 
are
tangent, where
and 
, 
the equations 
have exactly one common root, i.e.
have exactly one common root 
![$a\in\left\{\pm\frac {4\sqrt[4]{3}}{3}\right\}$](//latex.artofproblemsolving.com/b/e/9/be900390b4e7f11be9cc6ad45444a57cf705507d.png)
and in this case the unique real root (double) of the initial equation is ![$x_c=\left\{\begin{array}{ccc}
-\sqrt[4]3 & \mathrm{if} & a=\frac {4\sqrt[4]{3}}{3}\\\\
\sqrt[4]3 & \mathrm{if} & a=-\frac {4\sqrt[4]{3}}{3}\end{array}\right\|$](//latex.artofproblemsolving.com/c/f/5/cf5e9fc5ff3a8f4e42004d30e787d28da64b7e6f.png)
Proof 2. has exactly one real root has exactly one real root 
and verifies 
.
such that the polynomial 
can befactorized into the product of two polynomials with integer coefficients, positive degree. Then for such
factorize 
.
. Our polynomial admits the root 
or 
, i.e.
.
. Now appear two situations :
and in this case our polynomial becomes 
.
.PP2 (Uzbekistan NMO 2011). Find
such that 
is equal to the product of two non-constant polynomials with integer coefficients. . Our polynomial admits the root or , i.e. 
and in this case 
. . Now appear two situations :
. For 
our polynomial becomes 
.
For 
our polynomial becomes 
.PP3. Find
such that the polynomial equation 
has exactly one real root.Proof 1.
has exactly one real root 
has exactly one real root 
the graphs 
, 
aretangent, where
and 
, 
the equations 
have exactly one common root, i.e.
have exactly one common root 
![$a\in\left\{\pm\frac {4\sqrt[4]{3}}{3}\right\}$](http://latex.artofproblemsolving.com/b/e/9/be900390b4e7f11be9cc6ad45444a57cf705507d.png)
and in this case the unique real root (double) of the initial equation is ![$x_c=\left\{\begin{array}{ccc}
-\sqrt[4]3 & \mathrm{if} & a=\frac {4\sqrt[4]{3}}{3}\\\\
\sqrt[4]3 & \mathrm{if} & a=-\frac {4\sqrt[4]{3}}{3}\end{array}\right\|$](http://latex.artofproblemsolving.com/c/f/5/cf5e9fc5ff3a8f4e42004d30e787d28da64b7e6f.png)
Proof 2.
has exactly one real root has exactly one real root 
and verifies 
.This post has been edited 25 times. Last edited by Virgil Nicula, Nov 21, 2015, 6:33 PM
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