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

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For a polynomial P(x), there exists a polynomial Q(x) such that the product P(x)Q(x) results in a polynomial of x^3. The solution involves expressing P(x) in terms of its complex roots and using the identity related to roots of unity. Specifically, Q(x) can be constructed as the product of quadratic factors derived from the roots of P(x). This method is applicable to both complex and real polynomials, as real polynomials will have non-real roots in conjugate pairs, ensuring Q(x) remains a real polynomial. The discussion concludes by addressing a specific example to find Q(x) for a given P(x).
kaliprasad
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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 $x^3$
 
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kaliprasad said:
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 :
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
 
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Albert said:
my solution :
else $Q(x)=\dfrac {N(x^3)-R(x)}{P(x)}=M(x^2)$
so this $Q(x)$ always exists

in the else part the condition is not met.
 
kaliprasad said:
in the else part the condition is not met.
in this case $Q(x)=\dfrac {N(x^3)-R(x)}{P(x)}$
 
Albert said:
in this case $Q(x)=\dfrac {N(x^3)-R(x)}{P(x)}$

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

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
 
kaliprasad said:
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
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 $x^3$?
 
Albert said:
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
$x^4-x^2+1$ giving $x^6+1$
 
  • #10
kaliprasad said:
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]
 
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  • #11
answer for kaliprasad's question:
$P(x)=x^2+x+3$
find $Q(x)=?$
to make: $P(x)\times Q(x)=f(x^3)$
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"
 
  • #12
kaliprasad said:
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 $x^3$

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 $x^3$
 
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