Polynomial Division: Finding Q(x) for P(x)$x^3$

[kaliprasad]

Let P(x) be a polynomial of x. Show that there exists a polynomial Q(x) such that P(x)Q(x) is a polynomial of $x^3$

 

[Albert1]

Let P(x) be a polynomial of x. Show that there exists a polynomial Q(x) such that P(x)Q(x) is a polynomial of $x^3$

my solution :

Spoiler

let:
$P(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+------+a_0,(a_n\neq0)$ is a polynomial of $x$ with degree n
$M(x^2)=b_n(x^2)^n+b_{n-1}(x^2)^{n-1}+b_{n-2}(x^2)^{n-2}+------+b_0,\\
=b_n(x^{2n})+b_{n-1}(x^{2n-2})+------+b_0,(b_n\neq 0)$ is a polynomial of $x^2$ with degree n
$N(x^3)=c_n(x^3)^n+c_{n-1}(x^3)^{n-1}+c_{n-2}(x^3)^{n-2}+------+c_0,$
$=c_n(x^{3n})+c_{n-1}(x^{3n-3})+------+c_0,$ is a polymonial of $x^3$ with degree n
$(c_n\neq 0)$
set :$P(x)\times M(x^2)+R(x)=N(x^3)$ ,$R(x)$ is a polymonial of $x$ with degree less than n
if $R(x)=0 $ then $Q(x)=\dfrac {N(x^3)}{P(x)}=M(x^2)$
else $Q(x)=\dfrac {N(x^3)-R(x)}{P(x)}$
so this $Q(x)$ always exists

 

[kaliprasad]

my solution :

Spoiler

else $Q(x)=\dfrac {N(x^3)-R(x)}{P(x)}=M(x^2)$
so this $Q(x)$ always exists

Spoiler

in the else part the condition is not met.

 

[Albert1]

Spoiler

in the else part the condition is not met.

Spoiler

in this case $Q(x)=\dfrac {N(x^3)-R(x)}{P(x)}$

 

[kaliprasad]

Spoiler

in this case $Q(x)=\dfrac {N(x^3)-R(x)}{P(x)}$

Spoiler

but we want $N(x^3) = P(x)Q(x)$ which does not satisfy original requirement

 

[Albert1]

Spoiler

exam :
$x^3+1=(x+1)(x^2-x+1)+R(x), \,\, P(x)=x+1, Q(x)=x^2-x+1,R(x)=0$
$N(x^3)=x^3+1$ is a multiple of $x+1$
$x^3+2=(x+1)(x^2-x+1)+1---(A)$
the coefficient of $(A)\in Z$
$N(x^3)=x^3+2$ is not a multipl of $x+1$
but $x^3+2=x^3+(2^{\frac {1}{3}})^3=(x+2^{\frac {1}{3}})(x^2-2^{\frac {1}{3}}x+2^{\frac {2}{3}})---(B)$
here $P(x)=(x+2^{\frac {1}{3}}),Q(x)=(x^2-2^{\frac {1}{3}}x+2^{\frac {2}{3}})$
the coefficient of $(B)\in R$
the diversity occurs because the coefficients coming from different set

 

[kaliprasad]

Spoiler

exam :
$x^3+1=(x+1)(x^2-x+1)+R(x), \,\, P(x)=x+1, Q(x)=x^2-x+1,R(x)=0$
$N(x^3)=x^3+1$ is a multiple of $x+1$
$x^3+2=(x+1)(x^2-x+1)+1---(A)$
the coefficient of $(A)\in Z$
$N(x^3)=x^3+2$ is not a multipl of $x+1$
but $x^3+2=x^3+(2^{\frac {1}{3}})^3=(x+2^{\frac {1}{3}})(x^2-2^{\frac {1}{3}}x+2^{\frac {2}{3}})---(B)$
here $P(x)=(x+2^{\frac {1}{3}}),Q(x)=(x^2-2^{\frac {1}{3}}x+2^{\frac {2}{3}})$
the coefficient of $(B)\in R$
the diversity occurs because the coefficients coming from different set

Spoiler

if $P(x) = x^2 + 1$ then $ Q(x) = x^4 - x^2 - 1$
if $P(X) = x^2 +x + 1$ then $Q(x) = x - 1$
if $(P(X) = x^2 + x + 3$ then how do you find Q(x)
note: more than one Q(x) may be there and we are interested in anyone

 

[Albert1]

Spoiler

if $P(x) = x^2 + 1$ then $ Q(x) = x^4 - x^2 - 1$
if $P(X) = x^2 +x + 1$ then $Q(x) = x - 1$
if $(P(X) = x^2 + x + 3$ then how do you find Q(x)
note: more than one Q(x) may be there and we are interested in anyone

Spoiler

Do you consider $(x^2+1)(x^4-x^2-1)=x^6-2x^2-1=(x^3)^2-2(x^3)^\frac{2}{3}-1$ as a polynomial of $x^3$?

 

[kaliprasad]

Spoiler

Do you consider $(x^2+1)(x^4-x^2-1)=x^6-2x^2-1=(x^3)^2-2(x^3)^\frac{2}{3}-1$ as a polynomial of $x^3$?

My mistake I meant

Spoiler

$x^4-x^2+1$ giving $x^6+1$

 

[Opalg]

Let P(x) be a polynomial of x. Show that there exists a polynomial Q(x) such that P(x)Q(x) is a polynomial of $x^3$

[sp][This method only works if the coefficients are allowed to be complex numbers.]
Let $a_1,\ldots,a_n$ be the (complex) roots of $P(x)$, so that $$P(x) = c\prod_{j=1}^n(x-a_j)$$ for some constant $c.$ The identity $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$ shows that $$c\prod_{j=1}^n(x^3-a_j^3) = c\prod_{j=1}^n(x-a_j)(x^2 + a_jx + a_j^2).$$ So let $$Q(x) = \prod_{j=1}^n (x^2 + a_jx + a_j^2).$$ Then $$P(x)Q(x) = c\prod_{j=1}^n(x^3-a_j^3),$$ which is clearly a polynomial in $x^3.$

[/sp]
Edit:
[sp]In fact, that method also works for real polynomials. If $P(x)$ is a polynomial with real coefficients, then its non-real roots occur in adjoint pairs. Suppose that $x-a_j$, $x-a_k$ is a pair of adjoint factors (so $a_k$ is the complex conjugate of $a_j$). Then $x^2 + a_kx + a_k^2$ is the complex conjugate of $x^2 + a_jx + a_j^2$. So the product $(x^2 + a_jx + a_j^2)(x^2 + a_kx + a_k^2)$ will be real.

Thus any non-real factors of $Q(x)$ will also occur in complex conjugate pairs, and therefore $Q(x)$ will be a real polynomial.

[/sp]

 

[Albert1]

answer for kaliprasad's question:
$P(x)=x^2+x+3$
find $Q(x)=?$
to make: $P(x)\times Q(x)=f(x^3)$

Spoiler

let $Q(x)=x^4+ax^3+bx^2+cx+9$
$P(x)Q(x)=x^6+(a+1)x^5+(3+a+b)x^4+(3a+b+c)x^3+(9+3b+c)x^2+(9+3c)x+27=f(x^3)$
so we have $a=-1,b=-2,c=-3$
$Q(x)=x^4-x^3-2x^2-3x+9$
$P(x)Q(x)=x^6-8x^3+27$ is a polynomial of $x^3$
here $Q(x)$ is not unique,there are many other "possibilities"

 

[kaliprasad]

Let P(x) be a polynomial of x. Show that there exists a polynomial Q(x) such that P(x)Q(x) is a polynomial of $x^3$

Spoiler

we have
$P(x) = \sum_{n=0}^{\infty} a_nx^n$
this can be written as $A(x) + x B(x) + x^2C(x)$
where
$A(x) = \sum_{k=0}^{\infty} a_{3n}x^{3n}$
$B(x) = \sum_{k=0}^{\infty} a_{3n+1}x^{3n}$
$C(x) = \sum_{k=0}^{\infty} a_{3n+2}x^{3n}$
Now let
$R(x) = A(x) + x B(x)w + x^2C(x)w^2$ where w is cube root of one
and
$S(x) = A(x) + x B(x)w^2 + x^2C(x)w$
using $(a+b+c)(a+bw^2+cw)(a+bw+cw^2) = a^3+b^3+c^3 - 3abc$
we get $P(x)R(x)S(x) = (\sum_{k=0}^{\infty} a_{3n}x^{3n})^3 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^3 x^3 + (\sum_{k=0}^{\infty} a_{3n+2}x^{3n})^3x^6$
$- 3(\sum_{k=0}^{\infty} a_{3n}x^{3n})(\sum_{k=0}^{\infty} a_{3n+1}x^{3n})(\sum_{k=0}^{\infty} a_{3n+2}x^{3n})x^3$
which is a polynomial of $x^3$
now using $(a+bw+cw^2)( a+ bw^2 +cw) = (a^2+b^2+c^2-ab-bc-ca)$
we have
$R(x)S(x) = (A(x) + x B(x)w + x^2C(x)w^2)(A(x) + x B(x)w^2 + x^2C(x)w)$
$= (\sum_{k=0}^{\infty} a_{3n}x^{3n})^2 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^2)x^2 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^2x^4$
$- (\sum_{k=0}^{\infty} a_{3n+1}x^{3n}) (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})x$
$- (\sum_{k=0}^{\infty} a_{3n+1}x^{3n}))(\sum_{k=0}^{\infty} a_{3n+2}x^{3n})x^3$
$-(\sum_{k=0}^{\infty} a_{3n+2}x^{3n}\sum_{k=0}^{\infty} a_{3n}x^{3n})x^3$
multiplying by the above polynomial say Q(x) we get a polynomial of $x^3$