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(i) If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$ prove that:

$(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$

(ii)In this case prove that:

$ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor

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3 Answers 3

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Proving part (ii) independently is less tedious than part (i). Here it is:-

Let $f(x) = ax^5 + bx^2 + c$

Also $f(x) = (x^2 + px + 1)Q_3 (x)$; where $Q_3 (x)$ is a quotient polynomial of degree 3…….(1)

Let $g(x) = x^2 + px + 1$ such that g(x) = 0 has m and n as roots.

From product of roots, $n = 1/m$.

$f(n) = f(1/m) = … = (1/m)^5 . [a + b(m)^3 + c(m)^5]…....(2)$

From (1), $f(n) = f(1/m) = … = (1/m)^2 . [1 + p(m) + (m)^2]Q_3 (1/m)……(3)$

Equating (2) and (3), we have $a + bm^3 + cm^5 = (1 + pm + m^2).[m^3.Q_3 (1/m)]$

This means $a + bm^3 + cm^5$ has $(1 + pm + m^2)$ as factor.

i.e. $cx^5 + bx^3 + a$ has $(x^2 + px +1)$ as factor.

Thus, $ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor.

For part(i), I still need some time to workout.

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Part (i) is a continuation of Part (ii).

m is a root of g(x) = 0 and hence is a root of f(x) = 0.

Therefore, $am^5 + bm^2 + c = 0$……(4)

Using n = 1/m as root, from part (ii), we have $a + bm^3 +cm^5 = 0$…….(5)

Eliminating a from (4) & (5) by performing $(5)*m^5 – (4)$, we have

$bm^2(m^6 – 1) + c[m^T – 1] = 0$……(6) [T = 10 because of the display matter.]

Using the same technique to eliminate b and then c, we get

$a(m^6 – 1) – cm(m^4 – 1) = 0$……..(7)

$a(m^T – 1) + bm^3(m^4 – 1) = 0$ …...(8)

(6), (7), (8) combined yield the ratio of a : b : c such that, for some k

$a = –m^3(m^4 – 1) k$;

$b = (m^T – 1)k$; and

$c = –m^2(m^6 – 1)k$.

Using these values can verify that LHS = … = $m^6k^4(m^T – 1)^2(1 – m^4)^2$ = … = RHS.

Note-1 Who designs this question that involves so many calculations?

Note-2 This work is just doing the verification. How to get the required product is still unknown.

Note-3 Would like to know any other shorter method or better approach.

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@TomLynd I dug your question up from a long buried pile. Hope the proposed solution helps. –  Mick Oct 29 '13 at 18:08
    
@ Mick The question is from $Higher Algebra By JM Barnard Child$ –  Tom Lynd Oct 30 '13 at 16:44
    
@TomLynd Will definitely take a look at that book to see if there is any shortcut. –  Mick Oct 31 '13 at 4:50

This is a sequel to my previous work. The aim is to get the relation $(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$ directly rather than just verifying it. In order to do so, I have to re-arrange the original question as

“If $ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor $x^2+px+1$, then $(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$.”

[I can do that because the second part of the original question has been proved independently.]

For some polynomials P(x) and Q(x) of degree 3, we let $f(x) = ax^5 + bx^2 + c = (x^2 + px + 1)P(x)…………..(1)$ $g(x) = cx^5 + bx^3 + a = (x^2 + px + 1)Q(x)…………..(2)$

Let $m$ and $n$ (where $m.n = 1$) be the roots of $x^2+ px +1= 0$, then m is also a root of $f(x) = 0$ and of $g(x) = 0$. Thus,

$ am^5 + bm^2 + c = 0…..........(1.1)$

$ cm^5 + bm^3 + a = 0…......….(2.1)$

Do [a*(2.1) – c*(1.1)] and get $abm^3 – cbm^2 + (a^2 – c^2) = 0$

This means m is also a root of $F(x) = abx^3 – cbx^2 + (a^2 – c^2) = 0$ if we define

$F(x) = a*g(x) – c*f(x)$

On one hand $F(x) = abx^3 – cbx^2 + (a^2 – c^2)$

On the other hand $F(x) = (x^2+px+1).[aQ(x)] – (x^2+px+1)[cP(x)] = (x^2+ px+1).[R(x)$, a polynomial of degree 1 in x]

i.e. $F(x) = abx^3 – cbx^2 + (a^2 – c^2) = (x^2 + px + 1).[R(x)]$

Using the similar reasoning as before, $(a^2 – c^2)x^3 – cbx^1 + ab$ also has the same quadratic factor $(x^2+ px+1)$

For some polynomial, S(x) of degree 1 in x, define $G(x) = (a^2 – c^2)x^3 – cbx^1 + ab = (x^2 + px + 1).[S(x)]$

Since m is a root of $x^2 + px + 1 = 0$, m is also a root of $G(x) = (a^2 – c^2)x^3 – cbx^1 + ab = 0$

From $(a^2 – c^2)*F(x) – (ab)*G(x) = 0$, we have

$$x^2 – \frac {ab^2c} {bc(a^2 – c^2)}x + \frac {a^2b^2 – (a^2 – c^2)} {bc(a^2 – c^2)} = 0$$

The LHS of the above is actually that quadratic factor.

Hence, $[a^2b^2 – (a^2 – c^2)] / [bc(a^2 – c^2)] = 1$

The required relation in a, b and c then follows.

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@TomLynd I have the relation derived directly. –  Mick Nov 8 '13 at 2:49

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